He is actually my Stochastics tutor and i found this channel by accident. Im happy to see that more people than only the ones at our universty have the option to get his amazing explanations!
Me too :) I love how there's a bunch of higher math series on you-tube. I'm enjoying learning about the fundamental ideas without having to deal with too many picky details
@FriedIcecreamIsAReality I found some days ago that he is a teacher at TUHH. I can apply for the erasmus program there so I will try to find out more about him to see if I can possibly meet him
Amazing video! Your explanatory skills are quite something. Thank you so much for this series, your channel is a great contribution for Mathematics on the internet
6:22 All boxes are countable. There are uncountable quantity of boxes, but every box is countable. The obvious biyection over [x1], defined like y maps to y-x1 for every y in [x1] goes into Q. That is a countable set.
I really enjoy your videos, particularly that they focus on very important non trivial Math subjects, that are explained exceptionally well. Hope my professor of Measure Theory was nearly a fraction as good as you, maybe then I would not have become an algebraist ;) Keep it on! and Congratulations!
Thank you for your video. How can a normal student come up with such an equivalence partition and foresee a contradiction for this "standard exercise" ? (I struggled for a while to understand this)
i like to paraphrase arguments: lines are dense forests of points. if points had a measure bigger than zero, then lines would have an infinite measure, which we don't allow, so if you could ever measure a point or a line, their measures will be 0. (on the condition that you find a way to enumerate points countably, which he, in a sense, did) the conclusion is that we can't actually measure "points" bro, it's illegal and breaks everything.
I really recommend the video "How the Axiom of Choice Gives Sizeless Sets" from the PBS Infinite Series. They made some pretty and interesting visualizations of this paradox. Of course, I am not saying that in detriment of the brilliant (ha) work done in this video 😊.
Not everything is Lebesgue measurable? More like "Most definitely I bask in the multitude"...of facts that you share! Thanks for making such an amazing video series.
I am struggling in this question as well so I try to express what I think here in order to help anyone has the same question. Imagine A is the set that all the x ~ a and B is the set that all the x ~ b. If A n B has common elements c then c ~ a and c ~ b, for x ~ c it will be essentially x ~ a and x ~ b cause the transitivity of equivalent relationship. Then A and B must be the same.
I finally understood. However, I feel that the power set condition was a bit too strong than necessary for the proof. On a side note, I really feel like I'm having a personal tutor cuz you keep answering my questions LOL. I'm very grateful, cuz I understand immediately with your hints.
At 6:40 - 7:05, does this construction of A ("a set containing one and only one element from each of the distinct equivalence classes over (0,1]") depend on the Axiom of Choice? Because without AoC, it's not clear to me how A can be defined over an uncountably-large class of equivalence classes.. Edit: ok... you were getting to that in a minute!
C - א = C, the power of the irrational numbers. At 14:40+ the left inclusion in (c) is suspect, the union of the A_n being countably enumerated. What am I missing?
Hi! Thank you so much for the excellent measure theory resources, this course was never offered at my school so I’m really enjoying going through it myself this summer. Would you say this video series is best accompanied by a textbook? I don’t understand how I could have come up with this proof in a textbook given the material in the past episodes but I might just not be mathematically mature enough.
I have a small doubt regarding one of the steps of the proof. In one of the last steps the proof assumes that 0+0+0....=0. I have some reservation regarding this assumption. Doesnt the series evaluates to an indeterminate form ? If not, can you please elaborate on how is it safe to assume this? Thanks in advance. Regards Khushraj N M
As far as I understand when we sum exact zeros (0+0+..) we ofc get zero. The problem starts when we have limits. There we can get a number very close to 0, but not quite equal. We still write that it's 0 but in fact it's 0+delta (where delta is extremely small). And if we have (0+delta)*inf we get 0*inf (which is equal to 0 - the first part of my answer) + delta * inf. And latter is the problem as it can be anything. In the video we got that the measure is *exactly* zero, thus inf*0 = 0
The infinite sum 0+0+0+... is defined as the limit of the following sequence of partial sums: 0 0+0 0+0+0 ... In other words, the limit of the sequence 0,0,0,... which is 0.
Consider the cantor set. It has a measure of zero. Now, if you add all sets with exactly one middle segment, it as well has a measure of zero. Add all sets with exactly 2, 3,... n segments, each of those sets has a measure of zero. And those sets are disjoint. Yet, the sum of all sets (sets with zero, one, two,... infinitely many ones) add up to measure one.
Each of the equivalence classes are countable due to the countability of rational numbers, but the set (0, 1] itself is uncountable. So we can deduce that there are uncountably many eq classes
15:20 You assume that a measure is monotonic even though it's not specified in the measure problem nor in the definition of a measure, and you didn't prove it. So can a non-monotonic measure satisfying the measure problem exist?
Yeah the proof has to deal with considering continuous (regions) and discrete elements (points) of the power set to make a definition of a non-zero measure problematic. The continuous elements would be the segments (0,1) and (-1,2) in the video and the discrete elements are (A_1, A_2, ...)
So I think I understand this except for one bit. We were trying to prove that a measure on the power set of the reals didn't exist, but it looks like we instead proved the the only possible measure on the reals (not a power set of the reals) was the trivial measure. How do the two relate? Are the values in the interval (0,1] isomorphic to the power set of the reals?
The thing is that we take "representatives" (elements of A) to be from the interval (0,1]. Take a look at this one: ruclips.net/video/llnNaRzuvd4/видео.html
I have the same doubt. I think that the key idea is that he shows that the only possible measure for certain sets contained in the power set of reals was the trivial measure. Please correct me if you could solve this issue.
Power set is just the representation of all subsets. The measure problem is to measure all subsets of reals. And we construct a counter example which cannot be measured.
Hi, I noticed in your description that you say this explanation fits to lectures for students in their first year of study. Is it normal for first year students in Germany to study topics like the Lebesgue measure? Isn't that considered a more advanced part of analysis?
Hello! To be honest, "first year students" is in fact to optimistic. Most of students (mathematicians or physicists) learn the Lebesgue integral in the second year. However, you could consider it "advanced analysis" but sometimes lecturers skip the Riemann integral and introduce immediately the Lebesgue integral.
2:40 Why do you exclude the zero? The argument works fine for [0, 1] as far as I understand. Wikipedia article also doesn't mention any issues and works with [0, 1]: en.wikipedia.org/wiki/Vitali_set#Construction_and_proof
Hello! What a great series of videos! I just want to pinpoint something that I encounter every single time that greek letters are used in Math. μ is NOT pronounced as mu. It IS pronounced as "me".
Hello and thanks! The thing with Greek letters in Science and Maths is that different pronunciations are just common. I made a video about that: tbsom.de/s/ov The communication between mathematicians is different than the communication between Greeks. It's hard to fight against that :D
I'm a little confused by how the equivalence classes are disjoint (then again it's been awhile). I must be missing something key. If I recall correctly, [x] is all values of y that are related to x under the relation, right? So if I have x,y in I and x~y x-y in Q, then [x] = { y |x,y in I, x-y in I intersect Q } So if I pick x=0.1, then y can be all rational numbers from 0 exclusive to 0.1 exclusive because y=0.1 would give x-y < 0. So we have [0.1] = (0, 0.1) intersect Q But if I pick x=0.2, then y can be all rational numbers from 0 exclusive to 0.2 exclusive for the same reasons. So I have [0.2] = (0, 0.2) intersect Q But these are not disjoint. So me confuse! Please help!
I think your mistake is the assumption that your difference x-y in Q must be in I as well. The equivalence class [x] is all the numbers x+q that are in I while q is _any_ rational number. The value of q for x=0.1 will then go from -0.1 (actually just slightly bigger, since zero is excluded) to 0.9. And since 0.2 - 0.1 is a rational, those two numbers are in fact in the same equivalence class. In fact, I guess that every number you can write down in decimal form is in that same equivalence class, since you can always make a rational representation of the difference if there's a finite number of decimals? Uncountability is a bit confusing.
Now I started wondering whether it's true that the equivalence class [0.1] is in fact _exactly_ the set of those numbers in I that can be written in decimal form with a finite amount of ink…
Olli Wilkman No, some rational numbers are repeating decimals. 1/3 is in this same equivalence class but cannot be written in a finite number of digits, unless you are also including the notation of a bar over the digits to indicate repeating. But you are correct that the set of rational numbers in (0,1] is exactly one of these equivalence classes.
They are indeed disjoint. This is how i think of it: 1) if you had x=0.1 (or infact, any rational number), then the set [x] is essentially pretty much the set Q between 0 and 1 as [x] = { y |x,y in I, x-y in I intersect Q }, excluding 0 of course 2) therefore the numbers left, are irrational numbers. You cannot get from one irrational number to other certain irrational numbers easily via addition of rational numbers. You also cannot get from one irrational number and convert it back to a rational number through addition of other rational numbers. e.g. take x_2 = squareroot(0.1). now [x_2] will be some completely different disjoint set from the [x] made in one. 3) Repeat 2, with another irrational number not contained. Eventually, as there are infinitely many irrational numbers in 'I' (that is, (0,1]), there seems to be infinitely many disjoint sets. Visually, if you had a highlighter and highlight these numbers out on the number line (0,1], for any disjoint set would see infinitely many dots on the number line (0,1], so looks extremely similar visually, but will never overlap with another disjoint set.
Thanks all for the responses. Definitely the mistake I made was thinking y in I (as @Olli Wilkman said) which, if it was would give a bunch of subsets instead of a single subset of Q intersect I (in the case where x is rational). It comes back to not remembering everything about equivalence classes. I thought the relation being defined on I meant both variables x and y had to be on I also.
Of course we can develop measure theory in ZF, but as point out @The Bright Side of Mathematics, without choice we lost many important results in analysis (for instance, the Hahn-Banach theorem). However, there exists a weak statement called Dependent Choice (DC) and in ZF+DC we can get a lot of the results that are the crucial importance in analysis. Moreover, R. Solovay shows in the 70's (by forcing methods) that we can get a model of ZF+DC+every subset of the real line is measurable (and so, the axiom of choice doesn't hold in that model).
The intersection of An and Am should be a set of sets, each set is [xi], so I think we cannot directly get am - an = rn -rm because am and an are [xi] and we did not define the algebras on this set.
A quick question: why do we need to show that mu(R)=0 to show that mu=0 ? I was thinking (like for a map) hat we would have needed to show that mu(A)=0 for all A in the sigma-algrebra ? Or Am I missing something somewhere ?
Do I get this right (for the real number line), that one eq. class would contain ALL rational numbers (as each rational number can be reached from any other by adding or subtracting a rational number), and the other classes are basically only containing irrational numbers (one class per irrational number)? like [ sqrt(1/2) ] , [sqrt(5)] ,... etc. as you cant each one irrational number by r, which itself is rational
@@brightsideofmaths I assumed that the sum of irrational + rational numbers stay always irrational and that sqrt(5) = sqrt(1/2) + some_rational_number is not possible, which would result in that each irrational number would get its own equivalence class according to the definition given in your lecture. Please correct me if this is wrong. Why I'am asking this is just to get a better intuition for what the partition of R really means in this example
Not each irrational number gets its own equivalence class as you already mentioned: sqrt{5} and sqrt{5}+1/2 lie in the same equivalence class.@@martinschulze5399
@@brightsideofmaths By the way, thanks a lot for all the video content, its quite fun to learn new topics this way. I was actually studying probabilistic machine learning (I know most of the stuff actually but not in a stringent formal way, which I wanted to change now) ... until terms of measure theory popped up, where your lectures came in handy Uni Tubingen, Probl. ML ruclips.net/video/eN71go04iTc/видео.html
@@brightsideofmaths union An is countable because its formed from countable sets.... which is we can see that each equivalence classes posses exactly only one element
@@brightsideofmaths this leads to contradiction.... if m*(A)=0 then it will result m*(An)=0 for each n.... then sigma(m*(An))=0 too and this will implies to m*(0,1)=0... this is contradiction
If we can say that a_m - a_n is in Q means they are equivalent, why cannot we say all the enumeration of Q is just a single element? Is it more fair or does it break the proof? I feel like it really depends on how you treat a number.
@@shizhezang7511 I feel like you are getting confused between the two ways that (0,1] is being split up into disjoint sets. First, we split it up based on the equivalence relation to get the [x_i]'s.Each of these is one equivalence class, for example [1] would be the set of all rational numbers in (0,1]. another example is [1/sqrt2] which would be all the numbers that are a rational number away from 1/sqrt2. There are actually uncountably many of these equivalence classes, though this is not shown in the video, and each one has a countably infinite number of elements. Next we use this set of equivalence classes to come up with a NEW collection of disjoint sets, A_i's. To make the first A, just choose any element of EACH equivalence class (this will give you an uncountable number of elements). Then the rest of the A_i's are offset from this by a rational number. The A_i's are a countably infinite collection of sets, each with an uncountable number of elements- this is switched around from the [x_i]'s. So the enumeration of Q is ONE element of the collection of [x_i]s because it is one equivalence class. But it can also be used to keep track of all the A_i's because each element of [x_i] is in a different A_i. If A_n and A_m have a shared element, that means that they share the same choice from one of the [x_i]'s, so they must be the same set.
Use that every element y of (0,1] belongs to one of the equivalence classes and that the r_n's are an enumeration of all rational numbers in (-1,1]. Try this and you should be able to conclude that there is a in A and some r_n such that y=a+r_n, hence it is an element of the union of all A_n' s
@@brightsideofmaths I'd say we _will_ have uncountably many boxes. *Proof* (by contradiction): - Assume we have countably many boxes *[x_n]* - Notice each box has countably many elements, as *Q* is countable - Then *I* is the countable union of countable boxes => *I* is also countable Contradiction: *I* is uncountable (via binary representation + diagonal argument)!
real numbers make me uncomfortable. is there really not a way to talk about the ‘volume’ of a subset of the space of representable numbers, which are countable?
Interesting that you say we want to find the measure of length. Well, I don't, I'm here just for probability. Probability isn't fixed under translation...
But axiom of choice doesn't guarantee A_n and A_m are disjoint. For example: Let A = { {0}, {1}, {0,1}}. From axiom of choice, can choose A1={1}, A2={0}, A3={1}, while they are not disjoint. So the question is : how to guarantee or prove A_n and A_m are disjoint?
This is interesting, as it seems that we could avoid the contradiction if the measure of A were an infinitesimal surreal number, specifically something in [C,3C]/omega. That might open up a can of worms.
There are a lot of possibilities to do this. However, if we want to work in the usual mathematics foundations (with axiom of choice) and the real numbers, we have this measure problem here :D
I don't really understand your demonstration. You want to proove that there is no measure over the entire R. So, you claim that the measure mu is : Let a and b real numbers, A a subset of R, - mu([a, b]) = b - a - mu(x+A) = mu(A) And you want to proof that this don't exist. So you start with mu(]0, 1]) < infinity and mu(x+A) ) mu(A) and you want to proof that this is equal to zero. First, I don't understand why 0 and 1 and not just a and b. Then, you have a set I = ]0, 1] and you claim there is an equivalent relation on I such that: X, Y in I, X ~ Y X-Y in Q(the rational set) I don't undersant why because if I that pi/4 and 0.2 and I compute pi/4 - 0.2 it doesn't belong to Q. Then you define [x] like : [x] = {x+r | r in Q, x+r in I} You claim that all the [x_i] set are disjoints and why ? Because if I have [0] and [1/10] they are equals. I want to add that [pi/4] doesn't exist but pi/4 exists in I Then you take A is a subset of I and you claim: - For each [x], there is a "a" in A such that, "a" in [x] - For all a and b in A, a,b in [x] implies a = b. Why ? Assuming the properties, you define (M_n) n in N (natural set) is an enumeration of Q inter ]-1, 1] and you define A_n like A_n = M_n + A I don't understand why M_n ?
pi/4 and 0.2 are simply not equivalent. The first step would be to really understand this equivalence relation. I have a Start Learning Mathematics series where I explain a lot about equivalence relations.
why should I tasked herewith some annoying exercises? is this video to teach or to disturb me with some annoying tasks? In addition, even union of sets is badly defined here. In what sense union of An belongs to (-1,2] ?? This all makes it as worthy as any other SPAM here on youtube
@@brightsideofmaths u should tell from the start that u do not really provide the proof. Most are interested in raw material not in bumping into an exercises instead.
![BLACK BAG - Official Trailer [HD] - Only in Theaters March 14](http://i.ytimg.com/vi/Du0Xp8WX_7I/mqdefault.jpg)
He is actually my Stochastics tutor and i found this channel by accident.
Im happy to see that more people than only the ones at our universty have the option to get his amazing explanations!
Me too :) I love how there's a bunch of higher math series on you-tube. I'm enjoying learning about the fundamental ideas without having to deal with too many picky details
Lunatic108 ....Lucky you, he explains very well.
Which university?
@@JohnDoe927 yeah these videos are a saving grace, the notes given in class are pure heriogylfics
What is your university? I would like to have a professor like this too! The english sounds German or from norther Europe but I could be wrong
@FriedIcecreamIsAReality I found some days ago that he is a teacher at TUHH. I can apply for the erasmus program there so I will try to find out more about him to see if I can possibly meet him
I'm glad you showed the proof of this. There's no way I would have figured this out as an exercise left to the reader.
Amazing video! Amazing series! Please keep it coming! Measure theory has never been easier to understand. Thank you!!
I totally agree with you
Amazing video! Your explanatory skills are quite something. Thank you so much for this series, your channel is a great contribution for Mathematics on the internet
Thank you very much!
I like the way you demonstrated Vitali's proof. I look forward to more of your videos!
You are welcome! You are indeed right: That is a Vitali set and I wanted to elaborately demonstrate this proof. I hope that it helped :)
6:22 All boxes are countable. There are uncountable quantity of boxes, but every box is countable. The obvious biyection over [x1], defined like y maps to y-x1 for every y in [x1] goes into Q. That is a countable set.
This playlist still has weight now. Thank you very much, I understood so much about pure/classical mathematics :)
You're very welcome!
Wonderful videos! Thank you for sharing them!
....finally....
You managed to publish part 4 and part 5😊😊😊
Thanks a lot for these lectures! I am learning a lot since I discovered your channel. You explain things very well.
Thank you very much :)
I really enjoy your videos, particularly that they focus on very important non trivial Math subjects, that are explained exceptionally well. Hope my professor of Measure Theory was nearly a fraction as good as you, maybe then I would not have become an algebraist ;) Keep it on! and Congratulations!
Thank you for your video.
How can a normal student come up with such an equivalence partition and foresee a contradiction for this "standard exercise" ? (I struggled for a while to understand this)
i like to paraphrase arguments:
lines are dense forests of points. if points had a measure bigger than zero, then lines would have an infinite measure, which we don't allow, so if you could ever measure a point or a line, their measures will be 0.
(on the condition that you find a way to enumerate points countably, which he, in a sense, did)
the conclusion is that we can't actually measure "points" bro, it's illegal and breaks everything.
Excellent lecture!!
I really recommend the video "How the Axiom of Choice Gives Sizeless Sets" from the PBS Infinite Series. They made some pretty and interesting visualizations of this paradox. Of course, I am not saying that in detriment of the brilliant (ha) work done in this video 😊.
thanks, that video is like a perfect compliment of this video, it has nice visual but skip over the important proofs
Excellent explanation!
So grateful for these. Thanks so much.
Not everything is Lebesgue measurable? More like "Most definitely I bask in the multitude"...of facts that you share! Thanks for making such an amazing video series.
Thank you, you are the best
Can we get a quick link to an explanation for what are equivalence relations and classes? And also why those classes are disjoint?
Wikipedia is quite good there: en.wikipedia.org/wiki/Equivalence_class
if you're still unclear after a year you can ask! :)
I am struggling in this question as well so I try to express what I think here in order to help anyone has the same question. Imagine A is the set that all the x ~ a and B is the set that all the x ~ b. If A n B has common elements c then c ~ a and c ~ b, for x ~ c it will be essentially x ~ a and x ~ b cause the transitivity of equivalent relationship. Then A and B must be the same.
I finally understood. However, I feel that the power set condition was a bit too strong than necessary for the proof. On a side note, I really feel like I'm having a personal tutor cuz you keep answering my questions LOL. I'm very grateful, cuz I understand immediately with your hints.
Thank you for the videos.
Thank you very much! It's more than happy to now understand the knowledge that once troubled me a lot! Thank you again!
Thank you very much!
This is awesome. I still don't quite understand it but proofs are very hard for me.
Beautiful proof. pls make more videos!
At 6:40 - 7:05, does this construction of A ("a set containing one and only one element from each of the distinct equivalence classes over (0,1]") depend on the Axiom of Choice? Because without AoC, it's not clear to me how A can be defined over an uncountably-large class of equivalence classes..
Edit: ok... you were getting to that in a minute!
Totally correct!
You are a legend
C - א = C, the power of the irrational numbers.
At 14:40+ the left inclusion in (c) is suspect, the union of the A_n being countably enumerated.
What am I missing?
Which book do you use for the course. I am fluent in both german and english. Thank you.
Excellent! Thank you very very much!!
Glad it was helpful! :)
Hi! Thank you so much for the excellent measure theory resources, this course was never offered at my school so I’m really enjoying going through it myself this summer. Would you say this video series is best accompanied by a textbook? I don’t understand how I could have come up with this proof in a textbook given the material in the past episodes but I might just not be mathematically mature enough.
Thanks a lot! Definitely: you should complement this video course with a good textbook about the topic :)
thank you sir
Amazing
sir i am msc student .realy enjoy.thank you sir.
How have your studies been going?
fascinating, thank you.
how did mathematicians even construct this example to prove we cannot measure the power set of the real numbers in the first place, it's so deliberate
Mathematicians try a lot and sometimes have good ideas :D
Could you tell me which software are you using to write down the maths?
Xournal
Also called the Vitali set en.wikipedia.org/wiki/Vitali_set
Nice!
For the part c in the proof, the union of An should be within the interval of (-2,2] rather than (-1, 1]?
1:05 A measure problem : find a measure on P(R) which satisfies the desirable property
"You should try to prove it yourself"
Yes, why not? :)
would it be possible to know whats the software you use for whiteboard stuff?
egyptian student greet you , talented man
Thanks!
I have a small doubt regarding one of the steps of the proof. In one of the last steps the proof assumes that 0+0+0....=0. I have some reservation regarding this assumption. Doesnt the series evaluates to an indeterminate form ? If not, can you please elaborate on how is it safe to assume this? Thanks in advance.
Regards
Khushraj N M
For reference the assumption is made around 18:56.
@@khush625 This series is not an indeterminate form, because the infinite sum of 0 is countable (set of rational numbers)
As far as I understand when we sum exact zeros (0+0+..) we ofc get zero. The problem starts when we have limits. There we can get a number very close to 0, but not quite equal. We still write that it's 0 but in fact it's 0+delta (where delta is extremely small). And if we have (0+delta)*inf we get 0*inf (which is equal to 0 - the first part of my answer) + delta * inf. And latter is the problem as it can be anything.
In the video we got that the measure is *exactly* zero, thus inf*0 = 0
The infinite sum 0+0+0+... is defined as the limit of the following sequence of partial sums:
0
0+0
0+0+0
...
In other words, the limit of the sequence 0,0,0,... which is 0.
Consider the cantor set. It has a measure of zero. Now, if you add all sets with exactly one middle segment, it as well has a measure of zero. Add all sets with exactly 2, 3,... n segments, each of those sets has a measure of zero. And those sets are disjoint. Yet, the sum of all sets (sets with zero, one, two,... infinitely many ones) add up to measure one.
From the fact that the rational numbers are countable, can’t I assert that the set of all eq. classes {[x_n]} is countable as well?
Maybe you should think how many equivalence classes we actually have. And then you see if you can count them!
Each of the equivalence classes are countable due to the countability of rational numbers, but the set (0, 1] itself is uncountable. So we can deduce that there are uncountably many eq classes
15:20 You assume that a measure is monotonic even though it's not specified in the measure problem nor in the definition of a measure, and you didn't prove it. So can a non-monotonic measure satisfying the measure problem exist?
A measure is always monotonic and this follows from the properties with a short proof.
Yeah the proof has to deal with considering continuous (regions) and discrete elements (points) of the power set to make a definition of a non-zero measure problematic. The continuous elements would be the segments (0,1) and (-1,2) in the video and the discrete elements are (A_1, A_2, ...)
So I think I understand this except for one bit. We were trying to prove that a measure on the power set of the reals didn't exist, but it looks like we instead proved the the only possible measure on the reals (not a power set of the reals) was the trivial measure. How do the two relate? Are the values in the interval (0,1] isomorphic to the power set of the reals?
The thing is that we take "representatives" (elements of A) to be from the interval (0,1].
Take a look at this one: ruclips.net/video/llnNaRzuvd4/видео.html
I have the same doubt. I think that the key idea is that he shows that the only possible measure for certain sets contained in the power set of reals was the trivial measure. Please correct me if you could solve this issue.
Power set is just the representation of all subsets. The measure problem is to measure all subsets of reals. And we construct a counter example which cannot be measured.
I see why you chose the introduction of sigma algebra and measurable sets first.
Hi, I noticed in your description that you say this explanation fits to lectures for students in their first year of study. Is it normal for first year students in Germany to study topics like the Lebesgue measure? Isn't that considered a more advanced part of analysis?
Hello! To be honest, "first year students" is in fact to optimistic. Most of students (mathematicians or physicists) learn the Lebesgue integral in the second year. However, you could consider it "advanced analysis" but sometimes lecturers skip the Riemann integral and introduce immediately the Lebesgue integral.
@@brightsideofmaths Thanks for the reply:)
@@brightsideofmaths Gosh, Lebesgue integral is taught in my graduate course "real analysis". ....
@@yishi1022 That's what I said in the last sentence, isn't it? :) I don't think that it is a bad idea to that in the first year.
@@brightsideofmaths When you say first year, do you mean the in grad school or undergrad?
Nice video!
I am wondering is the axiom of choice neccesary here?
Yes, it is!
Sir can u plz provide solution PDF of that exercise.
It seems like that at 18:30 and 20:04, you had to use 0*\infty = 0.
No. I think you should google the definition of an infinite series. Under the normal definition, no “\infty” comes up.
Does the result change if we work in ZF without the axiom of choice?
The whole construction does not work then :D
You will have a hard to find such a non-measurable set.
2:40 Why do you exclude the zero? The argument works fine for [0, 1] as far as I understand. Wikipedia article also doesn't mention any issues and works with [0, 1]: en.wikipedia.org/wiki/Vitali_set#Construction_and_proof
Yes, it does not matter. I exclude zero to make the proof nicer later :)
@@brightsideofmaths I see, could you be more specific where it makes the proof nicer? Anyway, great series on the measure theory!
Later, we cover R with the half-open intervals.
Hello! What a great series of videos! I just want to pinpoint something that I encounter every single time that greek letters are used in Math. μ is NOT pronounced as mu. It IS pronounced as "me".
Hello and thanks! The thing with Greek letters in Science and Maths is that different pronunciations are just common. I made a video about that: tbsom.de/s/ov
The communication between mathematicians is different than the communication between Greeks. It's hard to fight against that :D
Wooooow! Just wow!
Thanks :)
I have a question that why your definition of An can get rational number set to be countable,they used to be uncountable.
I mean that rational number is dense even in the intervals of [-1,1], oh my teacher,you are my only light among the darkness.
@@HuangDuang Rational numbers are countable :)
11:35 put all the what on the other side??
Rs.
I'm a little confused by how the equivalence classes are disjoint (then again it's been awhile). I must be missing something key.
If I recall correctly, [x] is all values of y that are related to x under the relation, right? So if I have x,y in I and x~y x-y in Q, then [x] = { y |x,y in I, x-y in I intersect Q }
So if I pick x=0.1, then y can be all rational numbers from 0 exclusive to 0.1 exclusive because y=0.1 would give x-y < 0. So we have [0.1] = (0, 0.1) intersect Q
But if I pick x=0.2, then y can be all rational numbers from 0 exclusive to 0.2 exclusive for the same reasons. So I have [0.2] = (0, 0.2) intersect Q
But these are not disjoint. So me confuse! Please help!
I think your mistake is the assumption that your difference x-y in Q must be in I as well. The equivalence class [x] is all the numbers x+q that are in I while q is _any_ rational number. The value of q for x=0.1 will then go from -0.1 (actually just slightly bigger, since zero is excluded) to 0.9. And since 0.2 - 0.1 is a rational, those two numbers are in fact in the same equivalence class. In fact, I guess that every number you can write down in decimal form is in that same equivalence class, since you can always make a rational representation of the difference if there's a finite number of decimals? Uncountability is a bit confusing.
Now I started wondering whether it's true that the equivalence class [0.1] is in fact _exactly_ the set of those numbers in I that can be written in decimal form with a finite amount of ink…
Olli Wilkman No, some rational numbers are repeating decimals. 1/3 is in this same equivalence class but cannot be written in a finite number of digits, unless you are also including the notation of a bar over the digits to indicate repeating. But you are correct that the set of rational numbers in (0,1] is exactly one of these equivalence classes.
They are indeed disjoint. This is how i think of it:
1) if you had x=0.1 (or infact, any rational number), then the set [x] is essentially pretty much the set Q between 0 and 1 as [x] = { y |x,y in I, x-y in I intersect Q }, excluding 0 of course
2) therefore the numbers left, are irrational numbers. You cannot get from one irrational number to other certain irrational numbers easily via addition of rational numbers. You also cannot get from one irrational number and convert it back to a rational number through addition of other rational numbers.
e.g. take x_2 = squareroot(0.1). now [x_2] will be some completely different disjoint set from the [x] made in one.
3) Repeat 2, with another irrational number not contained. Eventually, as there are infinitely many irrational numbers in 'I' (that is, (0,1]), there seems to be infinitely many disjoint sets.
Visually, if you had a highlighter and highlight these numbers out on the number line (0,1], for any disjoint set would see infinitely many dots on the number line (0,1], so looks extremely similar visually, but will never overlap with another disjoint set.
Thanks all for the responses. Definitely the mistake I made was thinking y in I (as @Olli Wilkman said) which, if it was would give a bunch of subsets instead of a single subset of Q intersect I (in the case where x is rational).
It comes back to not remembering everything about equivalence classes. I thought the relation being defined on I meant both variables x and y had to be on I also.
Thanks a lot for this! How can I show that A is not a Borel set? Rather, how do Borel algebras solve this issue?
Why do we need the axiom of choice ?
Can we make a good measure of R in ZF ?
Yeah, without Axiom of Choice the measure theory gets easier. However, the axiom of choice is needed in other parts of analysis a lot.
Of course we can develop measure theory in ZF, but as point out @The Bright Side of Mathematics, without choice we lost many important results in analysis (for instance, the Hahn-Banach theorem). However, there exists a weak statement called Dependent Choice (DC) and in ZF+DC we can get a lot of the results that are the crucial importance in analysis. Moreover, R. Solovay shows in the 70's (by forcing methods) that we can get a model of ZF+DC+every subset of the real line is measurable (and so, the axiom of choice doesn't hold in that model).
The intersection of An and Am should be a set of sets, each set is [xi], so I think we cannot directly get am - an = rn -rm because am and an are [xi] and we did not define the algebras on this set.
Wrong
Wrong
Sorry, I have one question. When you establish A_n, you use a countable series. Is Q a countable set?
Yes.
rationals are countable
A quick question: why do we need to show that mu(R)=0 to show that mu=0 ? I was thinking (like for a map) hat we would have needed to show that mu(A)=0 for all A in the sigma-algrebra ? Or Am I missing something somewhere ?
You are missing the monotony of the measure.
@@brightsideofmaths Yes ! Thanks so much for your videos :)
Do I get this right (for the real number line), that one eq. class would contain ALL rational numbers (as each rational number can be reached from any other by adding or subtracting a rational number), and the other classes are basically only containing irrational numbers (one class per irrational number)? like [ sqrt(1/2) ] , [sqrt(5)] ,... etc. as you cant each one irrational number by r, which itself is rational
I think you mean the correct thing. The equivalence class [sqrt{1/2}] only contains irrational numbers.
@@brightsideofmaths I assumed that the sum of irrational + rational numbers stay always irrational and that sqrt(5) = sqrt(1/2) + some_rational_number is not possible, which would result in that each irrational number would get its own equivalence class according to the definition given in your lecture. Please correct me if this is wrong.
Why I'am asking this is just to get a better intuition for what the partition of R really means in this example
Not each irrational number gets its own equivalence class as you already mentioned: sqrt{5} and sqrt{5}+1/2 lie in the same equivalence class.@@martinschulze5399
@@brightsideofmaths oups, right! :) thanks!
@@brightsideofmaths By the way, thanks a lot for all the video content, its quite fun to learn new topics this way. I was actually studying probabilistic machine learning (I know most of the stuff actually but not in a stringent formal way, which I wanted to change now) ... until terms of measure theory popped up, where your lectures came in handy
Uni Tubingen, Probl. ML
ruclips.net/video/eN71go04iTc/видео.html
I do have a question why [x1] and [x2] cab be a disjointed set when x1 + r1 could possibly equal x2 + r2 if r2 were chosen to be x1-x2+r1
r2 has to be a rational number :)
♥️
Minutes 14:11 i still dont get it how could (0,1] be a subset of union An.... its still doesnt make sense... and i disbelieve this proof until now
Why does this not make sense?
@@brightsideofmaths union An is countable because its formed from countable sets.... which is we can see that each equivalence classes posses exactly only one element
@@matematikaextrim19mamankar4 However, An is not a countable set.
@@brightsideofmaths this leads to contradiction.... if m*(A)=0 then it will result m*(An)=0 for each n.... then sigma(m*(An))=0 too and this will implies to m*(0,1)=0... this is contradiction
You know what i mean? This contradicion will lead that the conclution that m* is not well defined in (0,1) . . This is the chaos
So beautiful
Please we need part 5,6,7,...
Part 5 is there :)
I feel like An = Am. However, An has no intersection on Am. Is this the reason that P(R) is not lebsgue meansurable?
If we can say that a_m - a_n is in Q means they are equivalent, why cannot we say all the enumeration of Q is just a single element? Is it more fair or does it break the proof? I feel like it really depends on how you treat a number.
@@shizhezang7511 I feel like you are getting confused between the two ways that (0,1] is being split up into disjoint sets. First, we split it up based on the equivalence relation to get the [x_i]'s.Each of these is one equivalence class, for example [1] would be the set of all rational numbers in (0,1]. another example is [1/sqrt2] which would be all the numbers that are a rational number away from 1/sqrt2. There are actually uncountably many of these equivalence classes, though this is not shown in the video, and each one has a countably infinite number of elements.
Next we use this set of equivalence classes to come up with a NEW collection of disjoint sets, A_i's. To make the first A, just choose any element of EACH equivalence class (this will give you an uncountable number of elements). Then the rest of the A_i's are offset from this by a rational number. The A_i's are a countably infinite collection of sets, each with an uncountable number of elements- this is switched around from the [x_i]'s.
So the enumeration of Q is ONE element of the collection of [x_i]s because it is one equivalence class. But it can also be used to keep track of all the A_i's because each element of [x_i] is in a different A_i. If A_n and A_m have a shared element, that means that they share the same choice from one of the [x_i]'s, so they must be the same set.
I have a hard time to show that (0, 1] should be included in the union of An. Could somebody help?
Use that every element y of (0,1] belongs to one of the equivalence classes and that the r_n's are an enumeration of all rational numbers in (-1,1]. Try this and you should be able to conclude that there is a in A and some r_n such that y=a+r_n, hence it is an element of the union of all A_n' s
why do we need the axiom of choice, if we already have a way to enumerate the x's can't we just take that A={x_1,x_2,x_3,......} ???
The enumeration was just a visualisation. We could have uncountably many boxes.
@@brightsideofmaths I'd say we _will_ have uncountably many boxes. *Proof* (by contradiction):
- Assume we have countably many boxes *[x_n]*
- Notice each box has countably many elements, as *Q* is countable
- Then *I* is the countable union of countable boxes => *I* is also countable
Contradiction: *I* is uncountable (via binary representation + diagonal argument)!
@@carstenmeyer7786 Yes, indeed :)
real numbers make me uncomfortable. is there really not a way to talk about the ‘volume’ of a subset of the space of representable numbers, which are countable?
Why mu(A)=0 given that the infinite sum is bounded but the convergence isn't necessarily zero?
Btw, Thanks for the video!
Saludos desde México
We sum up the same element. So there is only convergence if this element is zero :)
Interesting that you say we want to find the measure of length. Well, I don't, I'm here just for probability. Probability isn't fixed under translation...
For this, there is another video series :) See here: tbsom.de/s/pt
@@brightsideofmaths Thank you. I have already watched that one and came to the conclusion it isn't formal enough to help with my Probability classes.
It is formal but maybe not fitting for your class :)
One small question for the proof in Part 4: Why we need the fact that A_n and A_m are disjoint to show the following "subsets" relation in part (c).
:D, I have understood now.
But axiom of choice doesn't guarantee A_n and A_m are disjoint.
For example: Let A = { {0}, {1}, {0,1}}. From axiom of choice, can choose A1={1}, A2={0}, A3={1}, while they are not disjoint. So the question is : how to guarantee or prove A_n and A_m are disjoint?
@@marxman1010 I agree. What I have thought previously is that sigma-algebra requires the union of countably many disjoint sets.
@@danchenqijiang3008 Oh, I mistook one point that A is collection of disjoint sets. The example is not good, because those sets are not disjoint.
0*♾️ is indeterminate
But here, for measure theory, we define it.
This is interesting, as it seems that we could avoid the contradiction if the measure of A were an infinitesimal surreal number, specifically something in [C,3C]/omega. That might open up a can of worms.
There are a lot of possibilities to do this. However, if we want to work in the usual mathematics foundations (with axiom of choice) and the real numbers, we have this measure problem here :D
Could we omit this step ruclips.net/video/Ur3ofJ61bpk/видео.html and conlude n=m?
I don't really understand your demonstration.
You want to proove that there is no measure over the entire R. So, you claim that the measure mu is :
Let a and b real numbers, A a subset of R,
- mu([a, b]) = b - a
- mu(x+A) = mu(A)
And you want to proof that this don't exist.
So you start with mu(]0, 1]) < infinity and mu(x+A) ) mu(A) and you want to proof that this is equal to zero.
First, I don't understand why 0 and 1 and not just a and b.
Then, you have a set I = ]0, 1] and you claim there is an equivalent relation on I such that:
X, Y in I,
X ~ Y X-Y in Q(the rational set)
I don't undersant why because if I that pi/4 and 0.2 and I compute pi/4 - 0.2 it doesn't belong to Q.
Then you define [x] like :
[x] = {x+r | r in Q, x+r in I}
You claim that all the [x_i] set are disjoints and why ? Because if I have [0] and [1/10] they are equals.
I want to add that [pi/4] doesn't exist but pi/4 exists in I
Then you take A is a subset of I and you claim:
- For each [x], there is a "a" in A such that, "a" in [x]
- For all a and b in A, a,b in [x] implies a = b.
Why ?
Assuming the properties, you define (M_n) n in N (natural set) is an enumeration of Q inter ]-1, 1] and you define A_n like A_n = M_n + A
I don't understand why M_n ?
pi/4 and 0.2 are simply not equivalent. The first step would be to really understand this equivalence relation.
I have a Start Learning Mathematics series where I explain a lot about equivalence relations.
@@brightsideofmaths Yes this is why I don't understand your demonstration
@@brightsideofmaths I don't understand why you take those steps
@@Toto-cm5ux Do you know how to work with equivalence classes?
@@brightsideofmaths I think I forgot. Maybe I should see that first
why should I tasked herewith some annoying exercises? is this video to teach or to disturb me with some annoying tasks? In addition, even union of sets is badly defined here. In what sense union of An belongs to (-1,2] ?? This all makes it as worthy as any other SPAM here on youtube
No one forces you to watch this :D
@@brightsideofmaths u should tell from the start that u do not really provide the proof. Most are interested in raw material not in bumping into an exercises instead.
@@atenda-rg6hx I just guess you are joking.
The "Axiom of Choice" is the axiom of choice.
[see en.wikipedia.org/wiki/The_Treachery_of_Images]