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2d
comment It is possible to define our intuitive notion for probability in subsets of $[0,1]$
To make the argument completely rigorous, one has to show that the "winning-sets" of Ann and Bob are measurable in the product. This can be done and it actually holds that $2^{\omega_1}\otimes2^{\omega_1}=2^{\omega_1\times\omega_1}$, as shown here.
2d
comment It is possible to define our intuitive notion for probability in subsets of $[0,1]$
@Ilya Here is my favorite argument: Let $\omega_1$ be the first uncountable ordinal. Let Ann and Bob play the game of picking the larger ordinal number from $\omega_1$. Suppose there is a probability measure $\mu$ on $2^{\omega_1}$ that assigns measure $0$ to finite and therefore countable sets. If both play according to $\mu$, we get that for each $x\in\omega_1$, which is countable, Ann chooses a larger number with probability $1$. So by Fubini's theorem, Ann will win with probability $1$. By symmetry, so does Bob, which cannot be...
Oct
27
comment Minimax Theorems V.S. Fixed Point Theorems.
@Elias For the finite dimensional case, there is the nice book Fixed Point Theorems with Applications to Economics and Game Theory by Kim Border.
Sep
30
awarded  Explainer
Sep
26
comment Books for a beginner
@batpigandme I'm not familiar with the instructor's edition.
Sep
17
comment Can conditional expectation always be realized in a standard probability space?
@ElmarZander Yes, that is right.
Sep
4
awarded  Revival
Aug
31
comment Surprise exam paradox?
Yes, but that is compatible with the model.
Aug
31
comment Is it sufficient to compare two measures on a generator?
The way you asked the question, $\mathcal{G}=\emptyset$ would be the generator.
Aug
31
answered Is it sufficient to compare two measures on a generator?
Aug
25
comment Proof that $X$ countable, $\mathcal M$ algebra on $X$ implies $\mathcal M$ a $\sigma$-algebra
@MattBrenneman 1. You cannot accept comments as answers. I reposted the comment as an answer. 2. You can only accept the answer from the account from which you asked the question "Gremlin Brenneman".
Aug
25
answered Proof that $X$ countable, $\mathcal M$ algebra on $X$ implies $\mathcal M$ a $\sigma$-algebra
Aug
25
comment Proving that multiplication of convex function is convex
@roni It is false. Take the two functions $x\mapsto x$ and $x\mapsto -x$.
Aug
24
answered Metrizability of quotient spaces of metric spaces
Aug
24
comment Proof that $X$ countable, $\mathcal M$ algebra on $X$ implies $\mathcal M$ a $\sigma$-algebra
Yes, that's right.
Aug
24
comment Proof that $X$ countable, $\mathcal M$ algebra on $X$ implies $\mathcal M$ a $\sigma$-algebra
The family of all subsets of $\mathbb{N}$ that are either finite or have a finite complement is an algebra, but not a $\sigma$-algebra.
Aug
24
revised Reinventing The Wheel - Part 2: The Lebesgue Integral
added 13 characters in body
Aug
24
revised Reinventing The Wheel - Part 2: The Lebesgue Integral
added 10 characters in body
Aug
24
answered Reinventing The Wheel - Part 2: The Lebesgue Integral
Aug
23
answered “Structure of a measure space is the coarsest among all substantial structures on a set…”