YES YES YES REALLY NICE ONE. I was thinking "but can't instead I just f^-1(x) and solve for the same integral?" But when "f(x) = constant for x in a range" it's literally impossible to do that. And with this and the way you can apply it to probability where you can't just invert the function is brilliant!
As the King of Calculus, Riemann, faced the Dirichlet function, Riemann asked the function, “Are you Riemann integrable because you are continuous? Or are you continuous because you are Riemann integrable?” (note: this iff statement is not actually true) Riemann began to open his domain *”Malevolent Shrine”* and *cleaved* the area under the curve into slices approaching infinity. However, the Dirichlet function is continuous nowhere, and by *contradiction* the function simply stated “Stand proud. You're strong but Nah, I’d win.” Bernhard Riemann’s domain crumbled. In his dying moments, Riemann uttered the phrase “With this treasure, I summon…” Because in sets of measure zero always bet on *Lebesgue.* The proof was trivially concluded, and those who pioneered the techniques of calculus, the ones who formalised the integral, they would all bear witness to the one who is free. To the one who left it all behind, and his overwhelming *integrability!*
The day may come when you can download knowledge onto your brain like a computer. I'm not sure how universally jealous people would be of that though lol
I had to figure EVERYTHING out on my own!! I had a textbook and professors that tenure and were pissed about the number of courses they had to “teach”.
Yes and he should have written f(x) or something else, P(x) suggests probability but it's the density what he is plotting and is used to compute expectations. Amazing that the author hasn't pinned your comment yet
@@Pabloparsil Exactly but it remains a confusion between x the variable of integration and X the random variable. I would have written E[X] = integration of x *f(x) dx (f the density) or integration of x dP(x) with P the measure of the law of X (don't know of to say it properly sorry)
This is excellent. Every time I've tried to understand lebesgue integration in the past it was presented as something complicated and hard to understand, but you've made it so simple. I'm sure there's more detail you didn't cover, since once video can only do so much, but you've actually given some good intuition what what a lebesgue integral actual IS, which is something I haven't been able to find in any of the many other explanations I've found. Thank you very much!
One thing I'll give as an advice : don't stop. I looked back at myself from after 6 years and I'm ashamed of how better I was at 14/15/16/17. You're at an awesome level. Make sure it's not your peak! Awesome video :)
What field of maths did you learn at that age when you were younger so I can compare with myself. I feel like I'm learning too slowly to ever accomplish anything..
Mehmet M Mehmet M if that is the case that would be an unusual way of notating that. Usually the Probability = ∫ PDF(x)dx. In this case his P(x) is his PDF. The integral for E should be of E = ∫ x p(x) dx
This comment should be on the top pinned. I hate that the corrections always get buried under the fangirl comments. We get it... It's a great video. Can we get to scientific content distribution 2.0 already where this problem is sorted?
i follow both of the channels and man this feels very nice of everyone coming together and learning the topic so nicely. I really love the way you have presented lesbegue integral... I need to watch it more times for sure to get a little more understanding, but this is feels so nice! just switching how you construct rectangles to find area can do so much... would have never imagined that...
Where textbook lag; you proceed with your animation and make sense what the hell mathematics underneath which is very appealing to visualize absurd abstract idea from where concept are taken. Nice work bro; mathematics channel like you are life saver for viewer.
I think you haven't understood the Lebesgue integral at all. The animation with horizontal rectangles is quite flawed. That's not how it works!! Rectangles are vertical, what happens is that we split the range of the image into different sections. In each of them, we take an arbitrary point, and multiply it by the inverse function of the section. That may be one or more VERTICAL rectangles.
3:14 this is a bit too quick. While the probability for this one number is 0, we also have an infinite amount of other rational numbers it can be. So the total probability is an infinite sum of things. To determine that that is 0 you need some more arguments.
and it get paid out in the currency that somebody get some new knowledge, simply you get our satisfaction of knowing something, thank you, it is good, correct and sustainable.
The "limits of rectangles" explanation is what they teach you in high school calc, is how Newton/Leibniz thought of integration, and is correct to a good approximation, but it's not technically what a Riemann Integral is. Riemann Integrals (which are taught as a rigorous framework in Real Analysis class) are actually more powerful, and are able to integrate any function which is continuous "almost everywhere". ----------- Take Tomae's Function, which is extremely similar to Dirichlet's Function that you showed: f(x) = 0 if x is irrational, 1/n if x = m/n (rational) This function has infinite discontinuities. However, it's discontinuous at every rational number and continuous at every irrational number(!!). Since there are only countably-many discontinuities, it's continuous "almost everywhere" and is therefore Riemann Integrable! ----------- In fact, it was discovered in the 50's that a small tweak to the definition of Riemann Integrals makes them strictly stronger than Lebesgue Integrals - that is, every function which can be integrated using Lebesgue's Method can also be integrated using these new "Generalized Riemann Integrals", but the converse is not true!
@@BlueRaja *Riemann integrals are actually more powerful, and are able to integrate any function which is continuous "almost everywhere".* This is incorrect. There is only one Riemann integral, and the Riemann integral cannot integrate the indicator function of the rational numbers embedded in the real numbers, even though this function is bounded and continuous almost everywhere. In fact, this is one notable point of weakness of the Riemann integral that led to the development of the Lebesgue integral in the first place. Also, the claim that the Riemann integral is more powerful is just false. A well-known theorem of real analysis is that a function f : I -> R is Darboux integrable if and only if it is Riemann integrable. *Yes, that's the generalized Riemann integral.* Calling it this is misleading and incorrect, as this is not the name of the integral. Yes, the gauge integral is a direct generalization of the Riemann integral, but it is not called the generalized Riemann integral. Also, one important detail you neglect is that the gauge integral is only a stronger integral when we limit ourselves to functions in R^I. The measure-theoretic framework for the Lebesgue integral is more robust in that one can perform integration on arbitrary measurable spaces (X, Σ), rather than just (R, B(R)), which is the space in which the gauge integral is defined on. A more in-depth exploration of the subject reveals that the direction in which one generalizes from Riemann integration to measure-theoretic integration is a different one than the direction in which one generalizes from Riemann integration to gauge integration. It is possible to produce a tweak like the one from Riemann integration to gauge integration whilst simultaneously applying the Lebesgue framework of generalization of spaces. This shows that while the gauge integral is better behaved, it is not a broadening of scope in the same way Lebesgue's framework is.
@@angelmendez-rivera351 The "Generalized Riemann Integral" is the name we learned for it in Real Analysis class, and also the name my Real Analysis textbook uses ("Introduction to Real Analysis", Bartle+Sherbert, 1999), and is one of the names listed on the Wikipedia page. So I feel pretty confident in saying I'm not incorrect in calling it that. That same textbook also has this: Theorem 7.3.12 "Lebesgue's Integrability Criterion": A bounded function f : [a,b] → ℝ is Riemann Integrable if and only if it is continuous almost everywhere on [a,b] So I feel pretty confident about that claim, too. Your mistake is that "the indicator function of the rational numbers embedded in the reals", aka. Dirichlet's Function, is continuous nowhere (see en.wikipedia.org/wiki/Dirichlet_function#Topological_properties for a proof) --- When I said "Riemann Integrals are more powerful", I meant "more powerful than the limits of rectangles explanation would imply". Not "more powerful than Lebesgue integrals"
I really like the analogy to expected value. It's much more helpful than the coin analogy in Lebesgue's quote. The coin thing is too vague, but expected value is much more specific and so is a much more satisfying analogy. I also really appreciate that you gave a specific real world example of when Lebesgue integration would be useful. Personally, I particularly like the example with circuits as I'm an Electrical Engineering major.
@@vcubingx Thanks for making the video. 3b1b linked to it on his patreon and said it had similar animations to his videos. When I saw it was on Lebesgue integration I was excited, as every other explanation I'd found on the topic was too technical. A video similar in style to 3b1b that explained it sounded right up my alley and it was. I also watched your video on fractional calculus which was pretty interesting too.
The idea that somehow the Lebesgue integral is "done horizontally" is misleading. The Lebesgue integral is defined _exactly_ like the Riemann integral, except that for the Lebesgue integral the simple functions are linear combinations of characteristic functions of _measurable_ sets rather than just intervals. Then, once you've defined it, you prove that the integral is the area of the subgraph, you prove Fubini's theorem, and the "Cavalieri principle": all this allows you to rewrite the integral "horizontally" so to speak. No need to directly define the Lebesgue integral "horizontally".
Is the Graph at 6:50 correct? I'm not very deep into the mechanics, but as far as I understood the y_i * mu(A_i) is rectangle below the measurable Function similar to Rieman, but instead of equal x-length for the dx intervals the x-length is determined by dmu intervals on the y-axis. Please correct me if I am wrong.
@@vcubingx Maybe you can make another video on that topic. I think Tao's book did a good job of explaining it. (Chp 1.1 to 1.3 I think, I am learning this myself now too) An animated version of that will be fantastic!
Am I completely lost ? At 7:00 you say : _"consider trying to calculate the expected value of some probabilistic event if we're dealing with a probability distribution p of x like this normal distribution the expected value is just an integral from negative infinity to positive infinity of the distribution"_ But this is incorrect right ? Integrating the probability distribution (probability density for continuous variable) will simply give you 1 right ? This is one of Kolmogorov's probability axiom isn't it ? That the sum of the probabilities of all possible outcome in Ω be 1 ? In order to get the expected value, we need a weighted average, which means the integral of the value of the random variable, weighted by its probability.
Very nice explanation, it has been awhile since I took a measure theory class so my memory of it has diminished, but your explanation perfectly encapsulates the main concept behind Lebesgue integration.
This method is also the best way to integrate a normal distribution put into the 3rd dimension using multiplicative integrals but to do that it’s almost ALWAYS just 2 of them.
If you take an increasing sequence of Riemann integrable functions, the limit might not be Riemann integrable. And even if you happen to know the limit is Riemann integrable, it is still very tedious to prove that the limit of the integrals is the integral of the limit (a result, which seems obvious). With Lebesgue integrand this is almost trivial to prove. Historically this was very important. Another main motivation historically was Fubini's Theorem. There the Riemann integral is again in trouble with "integrability issues" . With Lebesgue integral Fubini is a breeze really.
At 7:38, the verbal emphasis on $5 is the same as that on $10, so (listening, not watching closely) I got confused in the subsequent math and had to back up to see what I'd missed. Not a complaint, just a note that inflection is relied upon by at least one viewer :). Thanks for the video!
I don't really like this "horizontal integral" analogy. That seems to me like Riemann but with a change of variables. You can consider a "partition" of the function's range with a similar intuition as in a cumulative distribution function, but the rectangles that are formed should not be horizontal, but still vertical. Lebesgue is about taking the inside vertical rectangles and the outside ones (taking the sup in the infinitesimal interval), showing their sum is essentially equal and then taking one of them as the integral, which turns out to be exactly as Riemann's (where the latter can be calculated). The fundamental change is in the fact that we do not care about which partition of an interval we take, but instead we consider all the simple functions (hence all possible values that f can reach, which to me is different from a partition of the range) and come to our result by density arguments. That's why it is called the "weak" integral, since proceeding by density we allow for functions that behave oddly in countable points and treat them as if they were normal, that's why Dirichlet's can be summed by Lebesgue. Of course this is an abstract construction in that you can't take all the simple functions when trying to calculate an integral. In fact numerical methods are based on Riemann's integral. Here is an article about why this horizontal visualization is actually misleading: en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics/Archive/2007/Jun#Mathematical_eyes_sought
Yes, it is the range of the function which is partitioned when dealing with Lebesgue integration. But the picture you draw in 3:40 is not what a simple function approximating the Lebesgue integral would look like and is slightly misleading.
Faulty reasoning, truth is we have no conceivable idea of what the 4th dimension would be, just like a 2D being can't conceive what the 3th dimension is, purely our assumptions that are probably wrong
The sources are awesome, I couldn't understand wikipedia but with "(Riemann) Integration Sucks!!!" from Peyam i could understand a lot more. Good video
Great job! I think the way you display the expected values around 8:14 is not the clearest,, namely because you have x_2 = -$5 \codt p_2 = 05. = -2.5 It took me a double take to realize that you're substituting p_2 = 0.5 into x_2 = -$5 \cdot p_2 and not have a single line equation. Some food for thought. The quote was brilliant: it helped me a ton.
Hey, nice video! Clear presentation 👌 If I may, can you skip less steps? Since it's a new concept for the viewer, it would help if you go slower even if you are using an argument you already went over earlier. It would help if you repeat the logic of the argument again while using it, just for clarity and connectedness. I just discovered your channel and Im excited to look at the rest of the videos! Keep goingg!! 👌👍
7:20 Giving/using for the random variable X the same symbol as the dummy integral variable x chosen after using the transformation theorem on the push-forward measure IP(X ∈ dx) = IP_X(dx) = (P o X^-1)(dx) to integrate with respect to the weighted Lebesgue measure p_X(x)dx involving the density p_X of X corresponding to the random variable X, is some kind of an abuse of notation ;). 7:26 Not of the distribution but an integral over the density p_X of the (absolutely continuous) random variable X. The distribution of a random variable is the image measure of the probability measure IP (on a probability space (Ω, A, IP)) under the random variable X. In the continuous case that would be exactly IP(X ≤ z) = 0∫z p_X(x) dx. 8:30 Here is, what I mean. On the left hand side there is a little x, but on the right hand side there is the random variable X.
Overall a good video, though I have a few comments. The figure featuring horizontal slabs on your thumbnail does not actually describe the Lebesgue integral. That figure indicates that we sum slabs of a fixed height and varying widths to approximate an integral. This is not accurate, as you describe yourself in your video. What happens instead is that we decompose the image of the function into a partition of intervals. The low end points of these intervals will be our weights for approximating an integral. We then collect the preimages of those intervals, which may be disconnected sets. However, these preimages will be measurable (by, e.g., continuity of the function. More generally by measurability of the function.) We then multiply the weights by measure of the preimages. In the coin counting analogy, the function values (which I'm calling weights) are the "number of coins of each type" and the measures of the preimages are the "values labeled on each coin." Also, I don't believe that the step function example is a good explanation as to why the Riemann integral is not preferable. There isn't any concern of mismatching upper & lower approximating sums in this case. Showing a visual of a more clear refinement demonstrates this. The problem with partitioning domains in high dimensional space is a good reason, but it isn't due to some conceptual complexity with domain decomposition. It is more that proofs on the properties of the multivariate Riemann integral require rather laborious notational baggage from multi-indexing high dimensional boxes combined with a time spent concerning the reader with the definition of measure zero sets while not actually trying to teach measure theory. Best reason to care about Lebesgue is going to be the use limiting properties, e.g. dominated convergence theorem.
Aren''t the integration components are still vertically distributed, because,it is still y*measure(or length) across x-axis? as per your diagram , it looks more like xdy instead of ydx, which is still a reimann integral with change in axis ?
Yes, you are correct. The claim that the Lebesgue integral can be intuitively visualized as being a vertical integral is just incorrect, and it leads to a false intuition of what the integral is. A much better intuitive explanation of the integral would remind us of the fact that there is no intrinsic reason to partition the interval of integration into closed intervals that serve as the base of rectangles, and that rearranging the points in those intervals without changing the value of the function at those points, intuitively, should not change the integral. Being able to partition the interval into arbitrary sets, instead of only other intervals, means that we need to have a well-defined notion of what the "length of a set" is, one which generalizes the intuitive idea of "length of an interval" that we already have. Lebesgue integration, and measure theory, are precisely one way to solve this problem.
The diagram is misleading in that it does not represent what is later described in the video. However, it is true that, for a non-negative real function f, there holds [Lebesgue integral of f against a finite measure m] = [Improper Riemann integral of m({f>=t})dt]. The diagram does represent the right-hand side. One advantage of the RHS is that you can replace m by a more general set function which is not a measure, in which case it is called a Choquet integral.
I learned the Darboux integral, not the Riemann integral. While they integrate precisely the same class of functions, they're conceptually different. With the Darboux integral, we consider the integrals of step functions less than or equal to the given function (the lower integral), and analogously for the upper integral. (Step functions are easy-peasy to integrate.) If the sup of the lower integrals equals the inf of the upper integrals, then that's the integral of the original function. (Tl;dr we approximate the function from above and below, using the method of exhaustion.) This generalizes to the Lebesgue integral; rather than approximate a function from below by a finite linear combination of characteristic functions of intervals, we substitute measurable sets for intervals. (Of course, now we need to define Lebesgue measure and Lebesgue-measurable sets.) To show that the rational numbers have measure 0, suppose we have epsilon paint, for some epsilon greater than 0. Since the rationals are denumerable, we order them r_1, r_2, r_3,... We use half our paint to cover r_1, one-fourth to cover r_2, one-eighth to cover r_3, etc. At the end, we've covered all the rationals with our paint, showing that they take up at most epsilon room. However, we can make epsilon as small as we like, so they take up less than any positive amount of room, and thus actually take up 0 room. (This vague concept of "taking up room" is what's meant by Lebesgue measure.)
@@subaruyagami2327 True, but I find them different conceptually. Specifically, it's very "hand-wavy" (imo) to claim that every Riemann sum of a function will converge to the same limit as the mesh size goes to 0. Of course, this can be made rigorous, but it seems to me that the Darboux integral is easier to establish, in that it is often simple to find partitions of an interval such that the associated upper and lower Darboux sums are within a given epsilon of the sought-after result, whereas showing the same result for every Riemann sum below a certain mesh size appears daunting. Again, this is based on my experiences and opinions, although I note that the Wikipedia entry on the Darboux integral states, "The definition of the Darboux integral has the advantage of being easier to apply in computations or proofs than that of the Riemann integral." (By no means the best-regarded source of information, but it's rather late for me and I'm headed to bed.)
What blows my mind is that the integral of the Dirichlet function is equal to 0, while the integral over the _negated_ Dirichlet function is equal to 1. The argument is the same, and I understand it in both cases, but still, it blows my mind.
I don’t like your 0.123123... It could be 0.12312345645612312345645600000...Rational but not matching your decimal string in the nth place. Still, you’re right of course that the integral is one, and if you switch to f(x)=1 if x is rational and 0 if x is irrational, the integral is 0. The rationals form an everywhere dense set of measure 0 on the interval.
2:40 How is the probability of getting 1 equal to 1/10 . There are 3 numbers it can be (1, 2, 3) and therefore it must be 1/3? And it was set up so as the decimal represents repeating 1,2,3 decimal, so why are we guessing the value of the next digit, when we know its 1, 2, 3?
The original idea of Lebesgue measure is the measure of the area under the graph which is later discovered to be coincided with sup of integrals of simple functions
If the only difference between the Riemann and Lebesgue integral was dividing up along the x-axis vs y-axis, then the Jordan measure (which uses boxes and doesn't care about the x or y axis) should be able to find the area under the graph of any Lebesgue-integrable function, right? The fact that this is not possible I think means there is some other deeper difference between the two integrals.
There are infinitely many decimal expansions starting at 0.123 that are rational, but in a precise sense, there are uncountably many more irrational numbers than rational numbers in a non-empty open interval; more than that, the rational numbers are only countably infinite, so by any reasonable notion of "measure", the set has measure zero, and that's why its contribution to the integral can be ignored (at least for Lebesgue and other measure-based integration definitions).
If it’s rational at some point it it has to repeat itself though, 123 was just an example. You can always say that at some point it has to stop and repeat itself otherwise it wouldn’t be rational.
This is true. I don't think he was trying to give a formal explanation though, so much as just trying to give a little bit of intuition. A slightly more formal version would invoke the concept of "countability" and of different sizes of infinity. I'm sure you can find a more extensive explanation online; however, in brief: we say that two sets have the "same size" or "same cardinality" if they can be put into a one-to-one correspondence (that is, there's a function from one set to the other set which is one-to-one and onto, and thus can be "inverted" or "undone" -- in fancy math language, this is called a "bijection"). This is indeed a reasonable definition: when you count something (say, a collection of sheep) you assign each sheep to a natural number in a strictly increasing way ("sheep number 1, sheep number 2, etc."), such that each sheep gets a number and no sheep is given multiple numbers (no double counting happens). You then see what subset of the natural numbers you're in correspondence with (if you've put the sheep in correspondence with the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, you say that there are 10 sheep; if you've put the sheep in correspondence with the set {1, 2, 3, ..., N}, you say that there are N sheep). The advantage of this definition, however, is that it allows you to compare the size of infinite sets. For instance, and slightly paradoxically, using this definition there are just as many positive even integers as there are positive integers: we can see this by the function from natural numbers to even natural number given by f(n) = 2n. In fact, this property is characteristic of infinite sets -- one definition of an infinite set is a set which can be put into one-to-one correspondence with a proper subset of itself. Using this definition, one can show that there are just as many rational numbers as there are positive integers, but that there are more real numbers than positive integers. To see this, let's introduce what's knows as the notion of a "height" of a rational number. The "height" of a rational number p/q (written in "simplest," or "reduced" form, with p and q sharing no common factors) is simply the larger of |p| or |q|, where the vertical bars denote absolute value, and can be thought of as sort of a measure of the "complexity" of a given rational number (so 1/2 is simpler than 47/355). Clearly, there's a finite number of rational numbers at any given height h, since there's finitely many integers between -h and h. Now, imagine listing the rational numbers in order of increasing heights (so, we write out all of the rational numbers with height 1 in some order, then all of the rational numbers with height 2 in some order, then all of the rational numbers with height 3 in some order, and so an and so on). Now that you've listed them, you've got a one-to-one correspondence: assign "1" to the first rational number on that list, "2" to the second, and so on and so on. Since every rational number has some height, every rational number appears somewhere on this list, and by construction appears only once, so this correspondence is indeed one-to-one. Therefor, we say that the "cardinality" of the rational numbers is the same as that of the positive integers, or that the two have the same "size." However, the same is not true of the real numbers. To see this, we use a "proof by contradiction." Imagine that the set of just the real numbers between 0 and 1 had the same size as the natural numbers. We can then imagine listing them out one after another in an infinite list, as we did with the rational numbers. We can also imagine writing out each real number as its infinite decimal expansion. Now, construct another number as follows: in the 1st decimal place of our new number, we write "7" if the 1st decimal place of the 1st number was NOT a 7, and if it was, we write "1". In the second decimal place of our new number, we write "7" if the 2nd decimal place of the 2nd number was NOT a 7, and if it was, we write "1." More generally, in the nth decimal place we write "7" if the nth decimal place of the nth number was NOT a 7, and if it was, we write "1." Now I claim this number is nowhere on our list: if it was somewhere on our list, then we could consider the number associated with the place it could be found on the list (that is, it would be at the mth place on this list for some value of m). But we know that the number we created disagrees with the mth number on our list in the mth decimal place--this yields a contradiction, since the number we created is clearly a real number, yet we had assumed that we'd listed all real numbers, and we've now created a real number which we failed to list! Since the set of real numbers includes those real numbers between 0 and 1, it certainly cannot be put into one-to-one correspondence with the positive integers, and since the set of real numbers includes all of the positive integers, the set of real numbers can be said to be the "larger" set. Now, we get to what vcubingx was saying -- because there are far far more real numbers than rational numbers, there must be far far more irrational numbers than rational numbers (since the set of reals is just the set of rationals combined with the set of irrational numbers). Thus, it is reasonable to say as he said in the video, that there are infinitely more real numbers than rational numbers. Of course, connecting this "cardinality" argument to something specific about "measure" requires more technical machinery than I've described here (in fact, more technical machinery than I yet have), but this long argument describes some of the intuition. Given its length, you can see why vcubingx didn't really want to get into all of this in the video.
@@kcsj9000 Yeah, precisely. Wasn't sure precisely how much background you had, so I wanted to do the argument out rather than simply use the analytic magic wand of saying "by Cantor's diagonal argument..."
@@nathanielkingsbury6355 From my long, rambling post above, you can show that any countable set has measure 0. Essentially, take epsilon paint, and cover the first point with half of it, the second with a quarter, the third with an eighth, etc. You paint the entire set with epsilon paint, which can be as small as you like; this is what is meant by having Lebesgue measure 0.
At 1:51 false Dirichlet function is integrated by rieman, and result is 0 But, of course, you can't understand this integral as sum of areas under the curve
I didn't get why the probability of .123123123.... being rational will be 0. From what I understand, the probability of choosing 0.123123.. from [0,1] is 0. Can you please explain this elaborately or give a link to where it is explained?
Another kind viewer gave a really good explanation to this, here's his comment if you can't find it: " This is true. I don't think he was trying to give a formal explanation though, so much as just trying to give a little bit of intuition. A slightly more formal version would invoke the concept of "countability" and of different sizes of infinity. I'm sure you can find a more extensive explanation online; however, in brief: we say that two sets have the "same size" or "same cardinality" if they can be put into a one-to-one correspondence (that is, there's a function from one set to the other set which is one-to-one and onto, and thus can be "inverted" or "undone" -- in fancy math language, this is called a "bijection"). This is indeed a reasonable definition: when you count something (say, a collection of sheep) you assign each sheep to a natural number in a strictly increasing way ("sheep number 1, sheep number 2, etc."), such that each sheep gets a number and no sheep is given multiple numbers (no double counting happens). You then see what subset of the natural numbers you're in correspondence with (if you've put the sheep in correspondence with the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, you say that there are 10 sheep; if you've put the sheep in correspondence with the set {1, 2, 3, ..., N}, you say that there are N sheep). The advantage of this definition, however, is that it allows you to compare the size of infinite sets. For instance, and slightly paradoxically, using this definition there are just as many positive even integers as there are positive integers: we can see this by the function from natural numbers to even natural number given by f(n) = 2n. In fact, this property is characteristic of infinite sets -- one definition of an infinite set is a set which can be put into one-to-one correspondence with a proper subset of itself. Using this definition, one can show that there are just as many rational numbers as there are positive integers, but that there are more real numbers than positive integers. To see this, let's introduce what's knows as the notion of a "height" of a rational number. The "height" of a rational number p/q (written in "simplest," or "reduced" form, with p and q sharing no common factors) is simply the larger of |p| or |q|, where the vertical bars denote absolute value, and can be thought of as sort of a measure of the "complexity" of a given rational number (so 1/2 is simpler than 47/355). Clearly, there's a finite number of rational numbers at any given height h, since there's finitely many integers between -h and h. Now, imagine listing the rational numbers in order of increasing heights (so, we write out all of the rational numbers with height 1 in some order, then all of the rational numbers with height 2 in some order, then all of the rational numbers with height 3 in some order, and so an and so on). Now that you've listed them, you've got a one-to-one correspondence: assign "1" to the first rational number on that list, "2" to the second, and so on and so on. Since every rational number has some height, every rational number appears somewhere on this list, and by construction appears only once, so this correspondence is indeed one-to-one. Therefor, we say that the "cardinality" of the rational numbers is the same as that of the positive integers, or that the two have the same "size." However, the same is not true of the real numbers. To see this, we use a "proof by contradiction." Imagine that the set of just the real numbers between 0 and 1 had the same size as the natural numbers. We can then imagine listing them out one after another in an infinite list, as we did with the rational numbers. We can also imagine writing out each real number as its infinite decimal expansion. Now, construct another number as follows: in the 1st decimal place of our new number, we write "7" if the 1st decimal place of the 1st number was NOT a 7, and if it was, we write "1". In the second decimal place of our new number, we write "7" if the 2nd decimal place of the 2nd number was NOT a 7, and if it was, we write "1." More generally, in the nth decimal place we write "7" if the nth decimal place of the nth number was NOT a 7, and if it was, we write "1." Now I claim this number is nowhere on our list: if it was somewhere on our list, then we could consider the number associated with the place it could be found on the list (that is, it would be at the mth place on this list for some value of m). But we know that the number we created disagrees with the mth number on our list in the mth decimal place--this yields a contradiction, since the number we created is clearly a real number, yet we had assumed that we'd listed all real numbers, and we've now created a real number which we failed to list! Since the set of real numbers includes those real numbers between 0 and 1, it certainly cannot be put into one-to-one correspondence with the positive integers, and since the set of real numbers includes all of the positive integers, the set of real numbers can be said to be the "larger" set. Now, we get to what vcubingx was saying -- because there are far far more real numbers than rational numbers, there must be far far more irrational numbers than rational numbers (since the set of reals is just the set of rationals combined with the set of irrational numbers). Thus, it is reasonable to say as he said in the video, that there are infinitely more real numbers than rational numbers. Of course, connecting this "cardinality" argument to something specific about "measure" requires more technical machinery than I've described here (in fact, more technical machinery than I yet have), but this long argument describes some of the intuition. Given its length, you can see why vcubingx didn't really want to get into all of this in the video. "
but how would you compute an integral like that. say f(x)=x^2. you need to find all the values where x^2>/=a so |x|>/=sqrt(a) => x from -sqrt(a) to sqrt(a). but then adding them up? seems very complicated. the area is S 2sqrt(p)*dp(can't make a greek p) but this is easy, x^2 is almost invertible what is we had something like sin(x)?
If u integrate a smooth function on a real interval, Lebesgue's integral is the same as Riemann's one. The difference shows up when you have more complicated functions on weird intervals
According to Riemann integration, f(x)=1 if x is rational, f(x)=0 if x is irrational, then the integral of f over [0,1] does not exist. But according to Lebesgue integration, the integral of f over [0,1] exists. Why? Can you explain clearly for me?
Your videos are great. Just an idea: Maybe it'd be nice to change a little bit more the font used in the videos to give them its own visual identity, right now they look too similar to 3b1b. Anyway, that's not a problem! Just something I think would be nice to the channel!
If its not possible to change the font then maybe it'd be a cool idea to choose a color pattern for the channel. Black background with blues graphs is the one 3b1b uses so maybe we could choose something else ! Again, just an idea I think would be beneficial to the channel.. love your content!
@@jorgemarcelo4708 Thanks for your idea, it's definitely something I'm gonna start doing. I should try using the same font which I use in my thumbnails, but I don't have any ideas for the background color. What do you think?
@@vcubingx In the website [colorhunt.co] it's possible to find lists of colors that go well together. It is even possible to search with tags like "pastel" colors. Maybe if you find a color pattern that you like the darkest color could be used in the background and the lighter ones in graphs.
@@jorgemarcelo4708 Thanks! I'm definitely gonna try these out. One of my friends also suggested [coolors.co] to generate some pallets, so I'll give both of these a try.
Are you planning to take real/complex analysis in high school? You're still in your junior year right? I think you'll have plenty of time! Btw, did 3b1b's probability series inspire you to do this video on Lebesgue integrals or was it just your own curiosity? Your videos are awesome, keep up the great work!
Nice one!
Thanks Grant!!
Acknowledged by the man, the myth, the legend.
He's here everybody! Shhhh! Get into your positions!
It warms my heart that Grant watched this video from a small youtuber
YES YES YES REALLY NICE ONE.
I was thinking "but can't instead I just f^-1(x) and solve for the same integral?"
But when "f(x) = constant for x in a range" it's literally impossible to do that.
And with this and the way you can apply it to probability where you can't just invert the function is brilliant!
As the King of Calculus, Riemann, faced the Dirichlet function, Riemann asked the function, “Are you Riemann integrable because you are continuous? Or are you continuous because you are Riemann integrable?” (note: this iff statement is not actually true) Riemann began to open his domain *”Malevolent Shrine”* and *cleaved* the area under the curve into slices approaching infinity. However, the Dirichlet function is continuous nowhere, and by *contradiction* the function simply stated “Stand proud. You're strong but Nah, I’d win.” Bernhard Riemann’s domain crumbled. In his dying moments, Riemann uttered the phrase “With this treasure, I summon…” Because in sets of measure zero always bet on *Lebesgue.* The proof was trivially concluded, and those who pioneered the techniques of calculus, the ones who formalised the integral, they would all bear witness to the one who is free. To the one who left it all behind, and his overwhelming *integrability!*
May be integrable in the sens of a distribution it seems like a dirac abstract distribution . I donow
This is beautiful! Did you write all of that yourself? Or is it an extract from somewhere?
@@tangpiseth8416 It's a brainrot copypasta from jujutsu kaisen edited for math 😅
So cool, more 3b1b inspired youtubers.. I am slightly jealous of kids today who are getting this level of education for almost free.
The day may come when you can download knowledge onto your brain like a computer. I'm not sure how universally jealous people would be of that though lol
one of the best things to happen to this world since 1940
I had to figure EVERYTHING out on my own!! I had a textbook and professors that tenure and were pissed about the number of courses they had to “teach”.
Yet many kids out there spend tons of hours watching TikTok dances
Yes, learning resources are more prevalent and easily accessible, but so are distractions imo
Clciked and thought it was 3b1b, glad I made that mistake, great video!
At 7:19, E(x) should be the integral of x*P(x), not just P(x). The integral of P(x) is 1!
Yes and he should have written f(x) or something else, P(x) suggests probability but it's the density what he is plotting and is used to compute expectations. Amazing that the author hasn't pinned your comment yet
Exactly right!
Yes
@@Pabloparsil Exactly but it remains a confusion between x the variable of integration and X the random variable. I would have written E[X] = integration of x *f(x) dx (f the density) or integration of x dP(x) with P the measure of the law of X (don't know of to say it properly sorry)
Great video Vivek! =)
Thanks Papa
Now I am waiting for some cursed meme about factoring f(x) and integrating Lebesgue Integeral with respect to miu.
Papa is here too omg
@@vcubingx is he your father?
@@Sciencedoneright his name is papa
This is excellent. Every time I've tried to understand lebesgue integration in the past it was presented as something complicated and hard to understand, but you've made it so simple. I'm sure there's more detail you didn't cover, since once video can only do so much, but you've actually given some good intuition what what a lebesgue integral actual IS, which is something I haven't been able to find in any of the many other explanations I've found. Thank you very much!
One thing I'll give as an advice : don't stop. I looked back at myself from after 6 years and I'm ashamed of how better I was at 14/15/16/17. You're at an awesome level. Make sure it's not your peak!
Awesome video :)
What field of maths did you learn at that age when you were younger so I can compare with myself. I feel like I'm learning too slowly to ever accomplish anything..
This is so cool for sure. Animated presentation is really good
7:25
Isn't it supposed to be the integral of
x p(x) dx
For the expected value function yes. The integral in the video sums to 1
I think it is P(X) = x*f(x), with f being the PDF.
Mehmet M Mehmet M if that is the case that would be an unusual way of notating that. Usually the Probability = ∫ PDF(x)dx. In this case his P(x) is his PDF. The integral for E should be of E = ∫ x p(x) dx
@@stefanjenkins6196 I know, that this was a typo. But nevertheless, I appreciate your nice explanation.
This comment should be on the top pinned. I hate that the corrections always get buried under the fangirl comments. We get it... It's a great video. Can we get to scientific content distribution 2.0 already where this problem is sorted?
Good video! You have an amazing talent of effectively illustrating complex ideas in a simple manner. Keep up the good work!!
Thank you very much!
This channel is gonna explode. Thanks for the explanation.
Mate I have an lesbegue integration exam in a few hours and was nearly having a panic attack till this video came along, thanks so much! Subscribed!
So much better than my Lebesgue Integration Course!!!
These animations just look so good :)
Thank you so much! Your videos were incredibly useful when I was learning this topic.
@@vcubingx Thanks. Glad I could help :)
@@vcubingx weak sauce.....not enough examples & theory....
i follow both of the channels and man this feels very nice of everyone coming together and learning the topic so nicely. I really love the way you have presented lesbegue integral... I need to watch it more times for sure to get a little more understanding, but this is feels so nice!
just switching how you construct rectangles to find area can do so much... would have never imagined that...
This channel deserves much more subscribers than it has now
Very clear and interesting explanation ! It was very cool to learn about the strengths of Lebesgue integral over Riemann integral. Thanks !
Where textbook lag; you proceed with your animation and make sense what the hell mathematics underneath which is very appealing to visualize absurd abstract idea from where concept are taken. Nice work bro; mathematics channel like you are life saver for viewer.
Superb!!!!!
This video just inspired me to extend this result into a deep area of number theory!
This channel has so nice quality! Congrats! :)
Thank you so much!
Your English has gotten very good!
Thanks, I'm very happy with my improvement too
I think you haven't understood the Lebesgue integral at all. The animation with horizontal rectangles is quite flawed. That's not how it works!! Rectangles are vertical, what happens is that we split the range of the image into different sections. In each of them, we take an arbitrary point, and multiply it by the inverse function of the section. That may be one or more VERTICAL rectangles.
3:14 this is a bit too quick. While the probability for this one number is 0, we also have an infinite amount of other rational numbers it can be. So the total probability is an infinite sum of things. To determine that that is 0 you need some more arguments.
Came to same this
and it get paid out in the currency that somebody get some new knowledge, simply you get our satisfaction of knowing something, thank you, it is good, correct and sustainable.
The "limits of rectangles" explanation is what they teach you in high school calc, is how Newton/Leibniz thought of integration, and is correct to a good approximation, but it's not technically what a Riemann Integral is. Riemann Integrals (which are taught as a rigorous framework in Real Analysis class) are actually more powerful, and are able to integrate any function which is continuous "almost everywhere".
-----------
Take Tomae's Function, which is extremely similar to Dirichlet's Function that you showed:
f(x) = 0 if x is irrational, 1/n if x = m/n (rational)
This function has infinite discontinuities. However, it's discontinuous at every rational number and continuous at every irrational number(!!). Since there are only countably-many discontinuities, it's continuous "almost everywhere" and is therefore Riemann Integrable!
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In fact, it was discovered in the 50's that a small tweak to the definition of Riemann Integrals makes them strictly stronger than Lebesgue Integrals - that is, every function which can be integrated using Lebesgue's Method can also be integrated using these new "Generalized Riemann Integrals", but the converse is not true!
damn, i shouldn't have taken up engineering.
This one, right?
en.wikipedia.org/wiki/Henstock-Kurzweil_integral
@@xCorvus7x Yes, that's the Generalized Riemann Integral.
@@BlueRaja *Riemann integrals are actually more powerful, and are able to integrate any function which is continuous "almost everywhere".*
This is incorrect. There is only one Riemann integral, and the Riemann integral cannot integrate the indicator function of the rational numbers embedded in the real numbers, even though this function is bounded and continuous almost everywhere. In fact, this is one notable point of weakness of the Riemann integral that led to the development of the Lebesgue integral in the first place. Also, the claim that the Riemann integral is more powerful is just false. A well-known theorem of real analysis is that a function f : I -> R is Darboux integrable if and only if it is Riemann integrable.
*Yes, that's the generalized Riemann integral.*
Calling it this is misleading and incorrect, as this is not the name of the integral. Yes, the gauge integral is a direct generalization of the Riemann integral, but it is not called the generalized Riemann integral. Also, one important detail you neglect is that the gauge integral is only a stronger integral when we limit ourselves to functions in R^I. The measure-theoretic framework for the Lebesgue integral is more robust in that one can perform integration on arbitrary measurable spaces (X, Σ), rather than just (R, B(R)), which is the space in which the gauge integral is defined on.
A more in-depth exploration of the subject reveals that the direction in which one generalizes from Riemann integration to measure-theoretic integration is a different one than the direction in which one generalizes from Riemann integration to gauge integration. It is possible to produce a tweak like the one from Riemann integration to gauge integration whilst simultaneously applying the Lebesgue framework of generalization of spaces. This shows that while the gauge integral is better behaved, it is not a broadening of scope in the same way Lebesgue's framework is.
@@angelmendez-rivera351 The "Generalized Riemann Integral" is the name we learned for it in Real Analysis class, and also the name my Real Analysis textbook uses ("Introduction to Real Analysis", Bartle+Sherbert, 1999), and is one of the names listed on the Wikipedia page. So I feel pretty confident in saying I'm not incorrect in calling it that.
That same textbook also has this:
Theorem 7.3.12 "Lebesgue's Integrability Criterion": A bounded function f : [a,b] → ℝ is Riemann Integrable if and only if it is continuous almost everywhere on [a,b]
So I feel pretty confident about that claim, too. Your mistake is that "the indicator function of the rational numbers embedded in the reals", aka. Dirichlet's Function, is continuous nowhere (see en.wikipedia.org/wiki/Dirichlet_function#Topological_properties for a proof)
---
When I said "Riemann Integrals are more powerful", I meant "more powerful than the limits of rectangles explanation would imply". Not "more powerful than Lebesgue integrals"
7:15 shouldn’t it be the integral of P(x)*x?
Can't wait to see how your content grows! Thanks for the video.
Thanks Aaron!
i keep seeing this vid's thumbnail and thinking it says "lesbian integrals". hats off for finding a way to stick out in my mind. youve earned my view
Thank you!!!!! you save me from failed in midterm!!!!
I really like the analogy to expected value. It's much more helpful than the coin analogy in Lebesgue's quote. The coin thing is too vague, but expected value is much more specific and so is a much more satisfying analogy.
I also really appreciate that you gave a specific real world example of when Lebesgue integration would be useful. Personally, I particularly like the example with circuits as I'm an Electrical Engineering major.
Glad it was helpful! Thanks for watching!
@@vcubingx Thanks for making the video. 3b1b linked to it on his patreon and said it had similar animations to his videos. When I saw it was on Lebesgue integration I was excited, as every other explanation I'd found on the topic was too technical. A video similar in style to 3b1b that explained it sounded right up my alley and it was. I also watched your video on fractional calculus which was pretty interesting too.
The idea that somehow the Lebesgue integral is "done horizontally" is misleading.
The Lebesgue integral is defined _exactly_ like the Riemann integral, except that for the Lebesgue integral the simple functions are linear combinations of characteristic functions of _measurable_ sets rather than just intervals.
Then, once you've defined it, you prove that the integral is the area of the subgraph, you prove Fubini's theorem, and the "Cavalieri principle": all this allows you to rewrite the integral "horizontally" so to speak.
No need to directly define the Lebesgue integral "horizontally".
Vivek my man !
Ur killing it with those animations !!! 🙌
Is the Graph at 6:50 correct? I'm not very deep into the mechanics, but as far as I understood the y_i * mu(A_i) is rectangle below the measurable Function similar to Rieman, but instead of equal x-length for the dx intervals the x-length is determined by dmu intervals on the y-axis. Please correct me if I am wrong.
This video is worth watching.... I really liked the way you described lebesgue integral really learnt something interesting subscribed ❤
1:01 Not to ruin what is an otherwise perfect video but continuity has a typo
Good catch! I'll see what I can do about it.
Yep. THere's an extra i the way he spelled it.
Ayanokoji learned this as a 5 year old
Reason why I came here bro Ayanokoji is just HIM.
I think the explanation will be even clearer if you mentioned more on what 'Lebesgue measure' is, at 5:07. Other than that great work!
I agree! Watching it back I realized that I skimmed over the idea of the Lebesgue measure, but it's definitely an incredibly important topic.
@@vcubingx Maybe you can make another video on that topic. I think Tao's book did a good job of explaining it. (Chp 1.1 to 1.3 I think, I am learning this myself now too) An animated version of that will be fantastic!
@@rzhang3927 I'll look into it! Definitely a fantastic idea.
Am I completely lost ?
At 7:00 you say : _"consider trying to calculate the expected value of some probabilistic event if we're dealing with a probability distribution p of x like this normal distribution the expected value is just an integral from negative infinity to positive infinity of the distribution"_
But this is incorrect right ? Integrating the probability distribution (probability density for continuous variable) will simply give you 1 right ? This is one of Kolmogorov's probability axiom isn't it ? That the sum of the probabilities of all possible outcome in Ω be 1 ?
In order to get the expected value, we need a weighted average, which means the integral of the value of the random variable, weighted by its probability.
Very nice explanation, it has been awhile since I took a measure theory class so my memory of it has diminished, but your explanation perfectly encapsulates the main concept behind Lebesgue integration.
This method is also the best way to integrate a normal distribution put into the 3rd dimension using multiplicative integrals but to do that it’s almost ALWAYS just 2 of them.
Great video! Lot of thanks for clarifying a frightening concept
If you take an increasing sequence of Riemann integrable functions, the limit might not be Riemann integrable. And even if you happen to know the limit is Riemann integrable, it is still very tedious to prove that the limit of the integrals is the integral of the limit (a result, which seems obvious). With Lebesgue integrand this is almost trivial to prove. Historically this was very important.
Another main motivation historically was Fubini's Theorem. There the Riemann integral is again in trouble with "integrability issues" . With Lebesgue integral Fubini is a breeze really.
Please do more of these vidoes! Good explanation. Would love some intuative measure theory / functional analysis.
At 7:38, the verbal emphasis on $5 is the same as that on $10, so (listening, not watching closely) I got confused in the subsequent math and had to back up to see what I'd missed. Not a complaint, just a note that inflection is relied upon by at least one viewer :). Thanks for the video!
I don't really like this "horizontal integral" analogy. That seems to me like Riemann but with a change of variables. You can consider a "partition" of the function's range with a similar intuition as in a cumulative distribution function, but the rectangles that are formed should not be horizontal, but still vertical.
Lebesgue is about taking the inside vertical rectangles and the outside ones (taking the sup in the infinitesimal interval), showing their sum is essentially equal and then taking one of them as the integral, which turns out to be exactly as Riemann's (where the latter can be calculated). The fundamental change is in the fact that we do not care about which partition of an interval we take, but instead we consider all the simple functions (hence all possible values that f can reach, which to me is different from a partition of the range) and come to our result by density arguments.
That's why it is called the "weak" integral, since proceeding by density we allow for functions that behave oddly in countable points and treat them as if they were normal, that's why Dirichlet's can be summed by Lebesgue.
Of course this is an abstract construction in that you can't take all the simple functions when trying to calculate an integral. In fact numerical methods are based on Riemann's integral.
Here is an article about why this horizontal visualization is actually misleading: en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics/Archive/2007/Jun#Mathematical_eyes_sought
I totally agree with you. The animation about integration using horizontal strips is very misleading, if not down right wrong.
You're talented my dude, keep it up.
Yes, it is the range of the function which is partitioned when dealing with Lebesgue integration. But the picture you draw in 3:40 is not what a simple function approximating the Lebesgue integral would look like and is slightly misleading.
To Visualize 4D object, think blue cube, split in half, and the second half is color coded sides that represent each cube value in a color.
Doesn't make sense
Faulty reasoning, truth is we have no conceivable idea of what the 4th dimension would be, just like a 2D being can't conceive what the 3th dimension is, purely our assumptions that are probably wrong
The sources are awesome, I couldn't understand wikipedia but with "(Riemann) Integration Sucks!!!" from Peyam i could understand a lot more. Good video
Great job! I think the way you display the expected values around 8:14 is not the clearest,, namely because you have
x_2 = -$5 \codt p_2 = 05. = -2.5
It took me a double take to realize that you're substituting p_2 = 0.5 into x_2 = -$5 \cdot p_2 and not have a single line equation.
Some food for thought. The quote was brilliant: it helped me a ton.
I didn't understand anything but the visuals are amazing. Great job!
Hey, nice video! Clear presentation 👌
If I may, can you skip less steps? Since it's a new concept for the viewer, it would help if you go slower even if you are using an argument you already went over earlier. It would help if you repeat the logic of the argument again while using it, just for clarity and connectedness.
I just discovered your channel and Im excited to look at the rest of the videos! Keep goingg!! 👌👍
Thanks for the feedback! I'll definitely keep this in mind when I make my next video!
7:20 Giving/using for the random variable X the same symbol as the dummy integral variable x chosen after using the transformation theorem on the push-forward measure IP(X ∈ dx) = IP_X(dx) = (P o X^-1)(dx) to integrate with respect to the weighted Lebesgue measure p_X(x)dx involving the density p_X of X corresponding to the random variable X, is some kind of an abuse of notation ;).
7:26 Not of the distribution but an integral over the density p_X of the (absolutely continuous) random variable X. The distribution of a random variable is the image measure of the probability measure IP (on a probability space (Ω, A, IP)) under the random variable X. In the continuous case that would be exactly IP(X ≤ z) = 0∫z p_X(x) dx.
8:30 Here is, what I mean. On the left hand side there is a little x, but on the right hand side there is the random variable X.
Overall a good video, though I have a few comments.
The figure featuring horizontal slabs on your thumbnail does not actually describe the Lebesgue integral. That figure indicates that we sum slabs of a fixed height and varying widths to approximate an integral. This is not accurate, as you describe yourself in your video. What happens instead is that we decompose the image of the function into a partition of intervals. The low end points of these intervals will be our weights for approximating an integral. We then collect the preimages of those intervals, which may be disconnected sets. However, these preimages will be measurable (by, e.g., continuity of the function. More generally by measurability of the function.) We then multiply the weights by measure of the preimages. In the coin counting analogy, the function values (which I'm calling weights) are the "number of coins of each type" and the measures of the preimages are the "values labeled on each coin."
Also, I don't believe that the step function example is a good explanation as to why the Riemann integral is not preferable. There isn't any concern of mismatching upper & lower approximating sums in this case. Showing a visual of a more clear refinement demonstrates this. The problem with partitioning domains in high dimensional space is a good reason, but it isn't due to some conceptual complexity with domain decomposition. It is more that proofs on the properties of the multivariate Riemann integral require rather laborious notational baggage from multi-indexing high dimensional boxes combined with a time spent concerning the reader with the definition of measure zero sets while not actually trying to teach measure theory. Best reason to care about Lebesgue is going to be the use limiting properties, e.g. dominated convergence theorem.
To say that It's a horisontal integral is not correct, because we need to calculate this sum:
sum k=1,n yk * M (yk
THANK YOU SO MUCH FOR THIS! Now I can learn the difference between classic reimann integration and this technique.
Glad it was helpful!
Dude, you are the god of animations. Lebesgue might be cool, but your animations are cooler.
Maybe, you animate Itô the next time. :)
Aren''t the integration components are still vertically distributed, because,it is still y*measure(or length) across x-axis? as per your diagram , it looks more like xdy instead of ydx, which is still a reimann integral with change in axis ?
Yes, you are correct. The claim that the Lebesgue integral can be intuitively visualized as being a vertical integral is just incorrect, and it leads to a false intuition of what the integral is. A much better intuitive explanation of the integral would remind us of the fact that there is no intrinsic reason to partition the interval of integration into closed intervals that serve as the base of rectangles, and that rearranging the points in those intervals without changing the value of the function at those points, intuitively, should not change the integral. Being able to partition the interval into arbitrary sets, instead of only other intervals, means that we need to have a well-defined notion of what the "length of a set" is, one which generalizes the intuitive idea of "length of an interval" that we already have. Lebesgue integration, and measure theory, are precisely one way to solve this problem.
The diagram is misleading in that it does not represent what is later described in the video. However, it is true that, for a non-negative real function f, there holds [Lebesgue integral of f against a finite measure m] = [Improper Riemann integral of m({f>=t})dt]. The diagram does represent the right-hand side.
One advantage of the RHS is that you can replace m by a more general set function which is not a measure, in which case it is called a Choquet integral.
You make such amazing videos 🙌
1:02 I don't see how this makes any sense. Either it works or it doesn't. You don't have to "visualize" anything.
I learned the Darboux integral, not the Riemann integral. While they integrate precisely the same class of functions, they're conceptually different. With the Darboux integral, we consider the integrals of step functions less than or equal to the given function (the lower integral), and analogously for the upper integral. (Step functions are easy-peasy to integrate.) If the sup of the lower integrals equals the inf of the upper integrals, then that's the integral of the original function. (Tl;dr we approximate the function from above and below, using the method of exhaustion.)
This generalizes to the Lebesgue integral; rather than approximate a function from below by a finite linear combination of characteristic functions of intervals, we substitute measurable sets for intervals. (Of course, now we need to define Lebesgue measure and Lebesgue-measurable sets.)
To show that the rational numbers have measure 0, suppose we have epsilon paint, for some epsilon greater than 0. Since the rationals are denumerable, we order them r_1, r_2, r_3,... We use half our paint to cover r_1, one-fourth to cover r_2, one-eighth to cover r_3, etc. At the end, we've covered all the rationals with our paint, showing that they take up at most epsilon room. However, we can make epsilon as small as we like, so they take up less than any positive amount of room, and thus actually take up 0 room. (This vague concept of "taking up room" is what's meant by Lebesgue measure.)
I also learned the Darboux integral instead of the Riemann integral in my undergrad analysis class.
@@tracyh5751 Were you using Apostol?
Riemann integral and Darboux integral are equivalent, so in a way you learned Riemann integral too!
@@subaruyagami2327 True, but I find them different conceptually. Specifically, it's very "hand-wavy" (imo) to claim that every Riemann sum of a function will converge to the same limit as the mesh size goes to 0. Of course, this can be made rigorous, but it seems to me that the Darboux integral is easier to establish, in that it is often simple to find partitions of an interval such that the associated upper and lower Darboux sums are within a given epsilon of the sought-after result, whereas showing the same result for every Riemann sum below a certain mesh size appears daunting. Again, this is based on my experiences and opinions, although I note that the Wikipedia entry on the Darboux integral states, "The definition of the Darboux integral has the advantage of being easier to apply in computations or proofs than that of the Riemann integral." (By no means the best-regarded source of information, but it's rather late for me and I'm headed to bed.)
@@tomkerruish2982 No, instructor's notes.
Nice , very nice . Please which App did you use for the graphic visualization of those area
What blows my mind is that the integral of the Dirichlet function is equal to 0, while the integral over the _negated_ Dirichlet function is equal to 1. The argument is the same, and I understand it in both cases, but still, it blows my mind.
7:05 There is an x missing under the integral? Otherwise the integral is just 1.
This is the best video 👍👍...Please sir give us more video about this topic and give us more example...
I don’t like your 0.123123... It could be 0.12312345645612312345645600000...Rational but not matching your decimal string in the nth place. Still, you’re right of course that the integral is one, and if you switch to f(x)=1 if x is rational and 0 if x is irrational, the integral is 0. The rationals form an everywhere dense set of measure 0 on the interval.
2:40
How is the probability of getting 1 equal to 1/10 . There are 3 numbers it can be (1, 2, 3) and therefore it must be 1/3? And it was set up so as the decimal represents repeating 1,2,3 decimal, so why are we guessing the value of the next digit, when we know its 1, 2, 3?
Beautiful explanation! Thanks so much for this video :)
Glad you enjoyed it!
The original idea of Lebesgue measure is the measure of the area under the graph which is later discovered to be coincided with sup of integrals of simple functions
can you make a step by step solution to some Lebesgue Integral?
I believe your expected value E[x] should be \int_{-\infty}^{\infty} x P(x) dx (see around 7:12)
This was amazing to watch 😊
Thanks, and it's so easy & simple!
If the only difference between the Riemann and Lebesgue integral was dividing up along the x-axis vs y-axis, then the Jordan measure (which uses boxes and doesn't care about the x or y axis) should be able to find the area under the graph of any Lebesgue-integrable function, right? The fact that this is not possible I think means there is some other deeper difference between the two integrals.
Nice explanation, will looking forward for other technique from you. 😀
Very clearly explained! Congrats!
Thanks a lot!
Thank you for this video! What about visualization of Ito Integral in contrast to Riemann/Lebesgue Integration please?
Glad you liked it! That's a tough topic to cover, so probably not anytime soon
Sir ! Which software did you use to make these awasome videos.
There are infinitely many decimal expansions starting at 0.123 that are rational, but in a precise sense, there are uncountably many more irrational numbers than rational numbers in a non-empty open interval; more than that, the rational numbers are only countably infinite, so by any reasonable notion of "measure", the set has measure zero, and that's why its contribution to the integral can be ignored (at least for Lebesgue and other measure-based integration definitions).
Great video mate
Thanks!
3:19 but there should be infinitely many possible rational numbers like 0.12311111... and 0.1231212121212...
If it’s rational at some point it it has to repeat itself though, 123 was just an example. You can always say that at some point it has to stop and repeat itself otherwise it wouldn’t be rational.
This is true. I don't think he was trying to give a formal explanation though, so much as just trying to give a little bit of intuition. A slightly more formal version would invoke the concept of "countability" and of different sizes of infinity. I'm sure you can find a more extensive explanation online; however, in brief: we say that two sets have the "same size" or "same cardinality" if they can be put into a one-to-one correspondence (that is, there's a function from one set to the other set which is one-to-one and onto, and thus can be "inverted" or "undone" -- in fancy math language, this is called a "bijection"). This is indeed a reasonable definition: when you count something (say, a collection of sheep) you assign each sheep to a natural number in a strictly increasing way ("sheep number 1, sheep number 2, etc."), such that each sheep gets a number and no sheep is given multiple numbers (no double counting happens). You then see what subset of the natural numbers you're in correspondence with (if you've put the sheep in correspondence with the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, you say that there are 10 sheep; if you've put the sheep in correspondence with the set {1, 2, 3, ..., N}, you say that there are N sheep). The advantage of this definition, however, is that it allows you to compare the size of infinite sets. For instance, and slightly paradoxically, using this definition there are just as many positive even integers as there are positive integers: we can see this by the function from natural numbers to even natural number given by f(n) = 2n. In fact, this property is characteristic of infinite sets -- one definition of an infinite set is a set which can be put into one-to-one correspondence with a proper subset of itself.
Using this definition, one can show that there are just as many rational numbers as there are positive integers, but that there are more real numbers than positive integers. To see this, let's introduce what's knows as the notion of a "height" of a rational number. The "height" of a rational number p/q (written in "simplest," or "reduced" form, with p and q sharing no common factors) is simply the larger of |p| or |q|, where the vertical bars denote absolute value, and can be thought of as sort of a measure of the "complexity" of a given rational number (so 1/2 is simpler than 47/355). Clearly, there's a finite number of rational numbers at any given height h, since there's finitely many integers between -h and h. Now, imagine listing the rational numbers in order of increasing heights (so, we write out all of the rational numbers with height 1 in some order, then all of the rational numbers with height 2 in some order, then all of the rational numbers with height 3 in some order, and so an and so on). Now that you've listed them, you've got a one-to-one correspondence: assign "1" to the first rational number on that list, "2" to the second, and so on and so on. Since every rational number has some height, every rational number appears somewhere on this list, and by construction appears only once, so this correspondence is indeed one-to-one. Therefor, we say that the "cardinality" of the rational numbers is the same as that of the positive integers, or that the two have the same "size."
However, the same is not true of the real numbers. To see this, we use a "proof by contradiction." Imagine that the set of just the real numbers between 0 and 1 had the same size as the natural numbers. We can then imagine listing them out one after another in an infinite list, as we did with the rational numbers. We can also imagine writing out each real number as its infinite decimal expansion. Now, construct another number as follows: in the 1st decimal place of our new number, we write "7" if the 1st decimal place of the 1st number was NOT a 7, and if it was, we write "1". In the second decimal place of our new number, we write "7" if the 2nd decimal place of the 2nd number was NOT a 7, and if it was, we write "1." More generally, in the nth decimal place we write "7" if the nth decimal place of the nth number was NOT a 7, and if it was, we write "1." Now I claim this number is nowhere on our list: if it was somewhere on our list, then we could consider the number associated with the place it could be found on the list (that is, it would be at the mth place on this list for some value of m). But we know that the number we created disagrees with the mth number on our list in the mth decimal place--this yields a contradiction, since the number we created is clearly a real number, yet we had assumed that we'd listed all real numbers, and we've now created a real number which we failed to list! Since the set of real numbers includes those real numbers between 0 and 1, it certainly cannot be put into one-to-one correspondence with the positive integers, and since the set of real numbers includes all of the positive integers, the set of real numbers can be said to be the "larger" set.
Now, we get to what vcubingx was saying -- because there are far far more real numbers than rational numbers, there must be far far more irrational numbers than rational numbers (since the set of reals is just the set of rationals combined with the set of irrational numbers). Thus, it is reasonable to say as he said in the video, that there are infinitely more real numbers than rational numbers. Of course, connecting this "cardinality" argument to something specific about "measure" requires more technical machinery than I've described here (in fact, more technical machinery than I yet have), but this long argument describes some of the intuition. Given its length, you can see why vcubingx didn't really want to get into all of this in the video.
@@nathanielkingsbury6355 diagonal argument?
@@kcsj9000 Yeah, precisely. Wasn't sure precisely how much background you had, so I wanted to do the argument out rather than simply use the analytic magic wand of saying "by Cantor's diagonal argument..."
@@nathanielkingsbury6355 From my long, rambling post above, you can show that any countable set has measure 0. Essentially, take epsilon paint, and cover the first point with half of it, the second with a quarter, the third with an eighth, etc. You paint the entire set with epsilon paint, which can be as small as you like; this is what is meant by having Lebesgue measure 0.
At 1:51 false
Dirichlet function is integrated by rieman, and result is 0
But, of course, you can't understand this integral as sum of areas under the curve
that function isnt riemann integrable because for any partion you take the upper reimann sum is 1 and the lower reimann sum is 0
2:44 Where does it come from?
I didn't get why the probability of .123123123.... being rational will be 0. From what I understand, the probability of choosing 0.123123.. from [0,1] is 0. Can you please explain this elaborately or give a link to where it is explained?
Another kind viewer gave a really good explanation to this, here's his comment if you can't find it:
"
This is true. I don't think he was trying to give a formal explanation though, so much as just trying to give a little bit of intuition. A slightly more formal version would invoke the concept of "countability" and of different sizes of infinity. I'm sure you can find a more extensive explanation online; however, in brief: we say that two sets have the "same size" or "same cardinality" if they can be put into a one-to-one correspondence (that is, there's a function from one set to the other set which is one-to-one and onto, and thus can be "inverted" or "undone" -- in fancy math language, this is called a "bijection"). This is indeed a reasonable definition: when you count something (say, a collection of sheep) you assign each sheep to a natural number in a strictly increasing way ("sheep number 1, sheep number 2, etc."), such that each sheep gets a number and no sheep is given multiple numbers (no double counting happens). You then see what subset of the natural numbers you're in correspondence with (if you've put the sheep in correspondence with the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, you say that there are 10 sheep; if you've put the sheep in correspondence with the set {1, 2, 3, ..., N}, you say that there are N sheep). The advantage of this definition, however, is that it allows you to compare the size of infinite sets. For instance, and slightly paradoxically, using this definition there are just as many positive even integers as there are positive integers: we can see this by the function from natural numbers to even natural number given by f(n) = 2n. In fact, this property is characteristic of infinite sets -- one definition of an infinite set is a set which can be put into one-to-one correspondence with a proper subset of itself.
Using this definition, one can show that there are just as many rational numbers as there are positive integers, but that there are more real numbers than positive integers. To see this, let's introduce what's knows as the notion of a "height" of a rational number. The "height" of a rational number p/q (written in "simplest," or "reduced" form, with p and q sharing no common factors) is simply the larger of |p| or |q|, where the vertical bars denote absolute value, and can be thought of as sort of a measure of the "complexity" of a given rational number (so 1/2 is simpler than 47/355). Clearly, there's a finite number of rational numbers at any given height h, since there's finitely many integers between -h and h. Now, imagine listing the rational numbers in order of increasing heights (so, we write out all of the rational numbers with height 1 in some order, then all of the rational numbers with height 2 in some order, then all of the rational numbers with height 3 in some order, and so an and so on). Now that you've listed them, you've got a one-to-one correspondence: assign "1" to the first rational number on that list, "2" to the second, and so on and so on. Since every rational number has some height, every rational number appears somewhere on this list, and by construction appears only once, so this correspondence is indeed one-to-one. Therefor, we say that the "cardinality" of the rational numbers is the same as that of the positive integers, or that the two have the same "size."
However, the same is not true of the real numbers. To see this, we use a "proof by contradiction." Imagine that the set of just the real numbers between 0 and 1 had the same size as the natural numbers. We can then imagine listing them out one after another in an infinite list, as we did with the rational numbers. We can also imagine writing out each real number as its infinite decimal expansion. Now, construct another number as follows: in the 1st decimal place of our new number, we write "7" if the 1st decimal place of the 1st number was NOT a 7, and if it was, we write "1". In the second decimal place of our new number, we write "7" if the 2nd decimal place of the 2nd number was NOT a 7, and if it was, we write "1." More generally, in the nth decimal place we write "7" if the nth decimal place of the nth number was NOT a 7, and if it was, we write "1." Now I claim this number is nowhere on our list: if it was somewhere on our list, then we could consider the number associated with the place it could be found on the list (that is, it would be at the mth place on this list for some value of m). But we know that the number we created disagrees with the mth number on our list in the mth decimal place--this yields a contradiction, since the number we created is clearly a real number, yet we had assumed that we'd listed all real numbers, and we've now created a real number which we failed to list! Since the set of real numbers includes those real numbers between 0 and 1, it certainly cannot be put into one-to-one correspondence with the positive integers, and since the set of real numbers includes all of the positive integers, the set of real numbers can be said to be the "larger" set.
Now, we get to what vcubingx was saying -- because there are far far more real numbers than rational numbers, there must be far far more irrational numbers than rational numbers (since the set of reals is just the set of rationals combined with the set of irrational numbers). Thus, it is reasonable to say as he said in the video, that there are infinitely more real numbers than rational numbers. Of course, connecting this "cardinality" argument to something specific about "measure" requires more technical machinery than I've described here (in fact, more technical machinery than I yet have), but this long argument describes some of the intuition. Given its length, you can see why vcubingx didn't really want to get into all of this in the video.
"
@@vcubingx ohh, I get the intuition behind it. Thanks! Great video, love it.
thankyou, that was a very clear explanation
Very nicely explained!!!
Wow. Until I heard his name said aloud, my brain automatically respelled Lebesgue's name as Lebesque even while reading it!
Excellent video lecture.
That bit he mentioned of fn tending towards f is called the Dominated Convergence Theorem I think
Indeed!
After looking at a couple of your videos I wondered, do you use 3b1b's manim python library for your animations?
I'm very sure he does... although you would have discovered it by now...
You are the 3Blue1Brown.Jr great video.... !!!
😂 I thought it was him when I clicked on the video
Do u mean lebesgue only improvised through the definition already given to the mathematical esperances in probabilities science 😊😊😊
but how would you compute an integral like that. say f(x)=x^2.
you need to find all the values where x^2>/=a so |x|>/=sqrt(a) => x from -sqrt(a) to sqrt(a). but then adding them up? seems very complicated.
the area is S 2sqrt(p)*dp(can't make a greek p) but this is easy, x^2 is almost invertible what is we had something like sin(x)?
If u integrate a smooth function on a real interval, Lebesgue's integral is the same as Riemann's one. The difference shows up when you have more complicated functions on weird intervals
7:18 x?
8:15 it looks as if for example x1 is equal to +10$ * p1 which is equal to 0.5 which is equal to 5. Please avoid these ambiguous equations.
Shouldn't the expectation be the integral of the probability distribution multiplied by x?
According to Riemann integration, f(x)=1 if x is rational, f(x)=0 if x is irrational, then the integral of f over [0,1] does not exist. But according to Lebesgue integration, the integral of f over [0,1] exists. Why? Can you explain clearly for me?
Cool.. Got to learn something new...
Please make a comparison video of reimann vs lebesgue integral, for the same calculation..
Your videos are great. Just an idea:
Maybe it'd be nice to change a little bit more the font used in the videos to give them its own visual identity, right now they look too similar to 3b1b.
Anyway, that's not a problem! Just something I think would be nice to the channel!
If its not possible to change the font then maybe it'd be a cool idea to choose a color pattern for the channel. Black background with blues graphs is the one 3b1b uses so maybe we could choose something else ! Again, just an idea I think would be beneficial to the channel.. love your content!
@@jorgemarcelo4708 Thanks for your idea, it's definitely something I'm gonna start doing. I should try using the same font which I use in my thumbnails, but I don't have any ideas for the background color. What do you think?
@@vcubingx In the website [colorhunt.co] it's possible to find lists of colors that go well together. It is even possible to search with tags like "pastel" colors. Maybe if you find a color pattern that you like the darkest color could be used in the background and the lighter ones in graphs.
I suggest typing "Winter" in the search bar, it'll give you "serious-looking" color palettes
@@jorgemarcelo4708 Thanks! I'm definitely gonna try these out. One of my friends also suggested [coolors.co] to generate some pallets, so I'll give both of these a try.
Are you planning to take real/complex analysis in high school? You're still in your junior year right? I think you'll have plenty of time! Btw, did 3b1b's probability series inspire you to do this video on Lebesgue integrals or was it just your own curiosity? Your videos are awesome, keep up the great work!
I don't think I will. The college I take my math courses at right now doesn't have an analysis course.
what program do you use