You are a wonderful person. I have been reading Billingsley's "Probability and Measure". Tbh, it's very enigmatic book and was struggling with it. Thank you for your clear explanations, your personal op ions on things, your reasoning. It makes measure theory so natural and simple because of you. Thank you!!!
I appreciate that you provide intuition for new definitions and gently introduce them. btw, "outer measure" is spelled with just one "t" (unless it's a Canadian spelling I'm unaware of).
At 1:14:25, why does the second equality follow from the first? While intuitive (the RHS is based upon a partition of E), its not clear to me why writing u*(E) in this manner is rigorously justified. We haven’t shown that u* is countably (or even finitely) additive, so why are we allowed break up u*(E) in this manner? Any help would be much appreciated!
Moreover, it appears to me that the first term on the RHS implies that the intersection of B1 and B2 is u*-measurable - the very thing we are trying to prove! This is because if we define E to be the ambient space (which we are allowed to do since E is allowed to be any subset of the ambient space), then the first term is equal to u* evaluated at the intersection of B1 and B2 - but a numerical value can only be returned if this argument is contained within the domain of u*, which is what we are trying to prove??😓
@27'45'' I think we can not let E be the entire set \Omega: as E \in \mathcal{A} and \mathcal{A} is just a ring not a field. And later @29'15'' E could be the entire of the countable union of the set of subsets {A_i}. Looking forward to hearing back from you. Thank you for the effort and quality of the lectures!
I think I have figured out the idea. For a pre-measure, it's defined on a semiring. For a outer measure, it is defined on the power set of $\Omega$ using caratheodory's extension with the pre-measure, where the usual convention $\inf \mathbb{\emptyset} = \infty$ applies. I found the above conclusion from the book "Infinite Dimensional Analysis: A Hitchhiker’s Guide 3ed".
It's quite confusing at the first time, as I spent a whole day long to find the answer. A lot of information I found about caratheodory theorem would skip the pre-measure and use the "algebra" rather than "semiring".
Looks like there is already a good answer here! But yes, the outer measure is defined on the power set, i.e. all subsets of Omega including Omega itself. This can be a bit confusing as introduced, because I was also discussing the idea of a premeasure and perhaps should have been a bit more careful in the exposition. But in Caratheodory's theorem, we extend the premeasure mu on ring A to the sigma-field generated by A. So in that case, there won't be an issue like this.
You are a wonderful person. I have been reading Billingsley's "Probability and Measure". Tbh, it's very enigmatic book and was struggling with it. Thank you for your clear explanations, your personal op ions on things, your reasoning. It makes measure theory so natural and simple because of you. Thank you!!!
I appreciate that you provide intuition for new definitions and gently introduce them.
btw, "outer measure" is spelled with just one "t" (unless it's a Canadian spelling I'm unaware of).
Underrated channel
I think for the inequality at 1:27:18 to hold we are assuming that the outer measure is continuous, which might still need some justification?
exactly why I'm browsing the comments as well
I figured it out, in case anyone else was confused. Monotonicity will suffice since B1^c cap B2^c cap ... Bn^c is a subset of B for every n.
At 1:14:25, why does the second equality follow from the first? While intuitive (the RHS is based upon a partition of E), its not clear to me why writing u*(E) in this manner is rigorously justified. We haven’t shown that u* is countably (or even finitely) additive, so why are we allowed break up u*(E) in this manner?
Any help would be much appreciated!
Moreover, it appears to me that the first term on the RHS implies that the intersection of B1 and B2 is u*-measurable - the very thing we are trying to prove! This is because if we define E to be the ambient space (which we are allowed to do since E is allowed to be any subset of the ambient space), then the first term is equal to u* evaluated at the intersection of B1 and B2 - but a numerical value can only be returned if this argument is contained within the domain of u*, which is what we are trying to prove??😓
@27'45'' I think we can not let E be the entire set \Omega: as E \in \mathcal{A} and \mathcal{A} is just a ring not a field. And later @29'15'' E could be the entire of the countable union of the set of subsets {A_i}. Looking forward to hearing back from you. Thank you for the effort and quality of the lectures!
I think I have figured out the idea. For a pre-measure, it's defined on a semiring. For a outer measure, it is defined on the power set of $\Omega$ using caratheodory's extension with the pre-measure, where the usual convention $\inf \mathbb{\emptyset} = \infty$ applies. I found the above conclusion from the book "Infinite Dimensional Analysis: A Hitchhiker’s Guide 3ed".
It's quite confusing at the first time, as I spent a whole day long to find the answer. A lot of information I found about caratheodory theorem would skip the pre-measure and use the "algebra" rather than "semiring".
Looks like there is already a good answer here! But yes, the outer measure is defined on the power set, i.e. all subsets of Omega including Omega itself. This can be a bit confusing as introduced, because I was also discussing the idea of a premeasure and perhaps should have been a bit more careful in the exposition. But in Caratheodory's theorem, we extend the premeasure mu on ring A to the sigma-field generated by A. So in that case, there won't be an issue like this.
sir i want your hand written notes
It's interesting. Did you recommend any book in order to follow all playlist???
At around 1:26, you say mu-star "less than" but you write "greater than." I think you meant what you said. Yes?
this is quite a burden on working memory, but very good none the less!
Il ilke the music in the beginning