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awarded  Good Answer
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awarded  probability
Sep
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comment A Vitali set is non-measurable, direct proof, without using countable additivity
Got it, thanks.
Sep
4
comment A Vitali set is non-measurable, direct proof, without using countable additivity
@AsafKaragila Can you construct Bernstein sets without the machinery of well-orderings, which areprobablynot familiar to someone just learning about Lebesgue measure?
Sep
4
comment A Vitali set is non-measurable, direct proof, without using countable additivity
How do you start the induction? In particular, how do you show that $m^*(V_F)\geq |F|m^*(V)$ when you only have subadditivity?
Sep
1
comment Borel isomorphism and approximation of Borel space valued function
@user138668 That's exactly it. The point is that you can take the topology that is most convenient to apply the arguments.
Aug
31
answered Borel isomorphism and approximation of Borel space valued function
Aug
30
answered Is every $\sigma$-algebra generated by a partition?
Aug
22
comment if i know that every Cauchy sequence converge then exists an unique $x\in \mathop{\cap}_{n=1}^\infty I_n $?
No. And the intervals have to be nested: $I_1\supseteq I_2\supseteq I_3\supseteq\ldots$
Aug
22
comment Prove a theorem involving normed spaces and completeness.
@user162343 If there is a linear homeomorphism, you can just use the easily proven fact that $\langle x_n\rangle$ is Cauchy if for every neihborhood $O$ of $0$, there is $N$ such that $x_n-x_m\in O$ for all $n,m\geq N$.
Jul
10
comment Showing that $[0,1]$ is compact
@StanShunpike That is basically the definition of an open set. If $O_s$ is open, there must be some "wiggle room" around each of its elements. In particular, if $s\in O_s$ then for some $\epsilon>0$, $(s-\epsilon, s+\epsilon)\subseteq O_s$. The result follows now from $(s-\epsilon,s]\subseteq (s-\epsilon,s+\epsilon)\subseteq O_s$.
Jun
14
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May
31
comment Advantage of accepting non-measurable sets
@RonMaimon I'm looking forward to seeing your definition of "arbitrary shape".
May
30
comment Signaling game : response to zero-probability message
@Guilro This game has no proper subgame, if you take account of the choice of nature.
May
10
comment Is Riesz measure an extension of product measure?
I guess $f$ should be in $C(X\times Y)$.
Apr
3
answered Axioms Conditional Probability
Apr
1
answered Decomposition of measures as a product measures