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Jul
3
revised Is this space normal?
edited the topic title and one sentence.
Jul
3
suggested approved edit on Is this space normal?
Jun
29
comment Question about proving something using reverse Fatou's lemma
@madprob: You're right, thanks. And even though I wrote it there, I actually didn't use it in the proof. Indeed $\infty-\infty$ can never occur there as $\mu(g)<\infty$.
Jun
29
revised Question about proving something using reverse Fatou's lemma
took the integrability of $f_{n}$ out.
Jun
28
comment bounded measurable function is the uniform limit of a sequence of simple functions
The key point behind the uniform part of this statement lies in the boundedness of $f$.
Jun
19
comment Difference between supremum and maximum
@copper.hat: The two sentences seem to contradict each other. Correct me if I'm wrong, but first you say maximum is attained and supremum might not, and then you give an example where supremum is attained but maximum isn't. Maybe I misunderstood your point?
Jun
19
comment Subset of $\mathbb{I}\cap [0,1]$ (irrationals in [0,1]) that is closed in $\mathbb{R}$ and has measure $\epsilon \in (0,1)$
In what text did you see this and what page?
Jun
19
comment A simple question about open set
@Mathematics $B(-1,r)=\{x\in S:d(x,-1)<r\}$ as a nhood of $-1$ in $S$ is by definition a subset of $S$. Ofcourse the open ball in $X$ is not a subset of $S$.
Jun
18
comment Continuous images of open sets are Borel?
Thanks Brian, this answer is also very enlightening.
Jun
18
awarded  Nice Question
Jun
18
comment Continuous images of open sets are Borel?
@MichaelGreinecker: Thanks!
Jun
18
comment Continuous images of open sets are Borel?
@t.b. True, that would do the job. Thanks.
Jun
18
accepted Continuous images of open sets are Borel?
Jun
18
asked Continuous images of open sets are Borel?
Jun
17
comment Compactness in $\mathbb{Q}$
@MarcoCastronovo: You can do it the following way. Let $A\subset X$ and give $A$ the subspace topology of $X$. Show that $A$ is compact if and only if every cover of $A$ in $X$ has a finite subcover. The role of $X$ was arbitrary, as long as the topology of $A$ is the subspace topology from $X$.
Jun
17
comment Sigma-field of a sequence of Random Variables
Okey. Just as I had assumed too.
Jun
17
comment Sigma-field of a sequence of Random Variables
By the way, it is different from saying that we have a random variable $\tilde{X}:\Omega\to\mathbb{R}^{\infty}$ given by $\tilde{X}(\omega)=(X_{1}(\omega),X_{2}(\omega),...)$ than saying that $\tilde{X}=(X_{i})_{i=1}^{\infty}$ is a sequence of random variables. When looking at the $\sigma$-algebras they generate, in first case you need to consider preimages of Borel sets in $\mathbb{R}^{\infty}$ and in latter you need to consider preimages or Borel sets in $\mathbb{R}$ for each $X_{n}$. And whether these two agree depends on the topology of $\mathbb{R}^{\infty}$. Which one did you mean?
Jun
17
revised Sigma-field of a sequence of Random Variables
added 120 characters in body
Jun
17
answered Sigma-field of a sequence of Random Variables
Jun
16
comment Form of $\sigma(X_n)$ and $\sigma(X_n)$-measurability
@Justin: What I meant was, that $\sigma(\tilde{X})=\sigma(\{\{X_{n}\leq a\}:a\in\mathbb{R},\,\,n\in\mathbb{N}\})$, i.e. that $\{X_{n}\leq a\}$ form of sets are generators of $\sigma(\tilde{X})$. This is straight from the definition, as $\sigma(\tilde{X})$ is the smallest $\sigma$-algebra for which respect all the functions $X_{n}$ are measurable, and $X_{n}$ to be measurable is equivalent of saying that $\{X_{n}\leq a\}$ is a measurable set for all $a\in\mathbb{R}$ since $\{]-\infty,a]:a\in\mathbb{R}\}$ generates the Borel algebra of $\mathbb{R}$.