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20h
comment Is there a name for functions $f$ such that $\{f \le a\}$ is analytic?
@PedroSánchezTerraf: Interesting, I will have to look up "grey sets". Any suggested references? Google gives lots of, shall we say, irrelevant hits. I guess partially ordered Polish spaces could be interesting also but I was thinking of the lexicographic order on $\omega^\omega$. Not really for any good reason.
1d
comment If ${a_i} \to 0$ and $\{ {X_i}\} _{i = 1}^\infty $ is a sequence of iid random variables with zero mean, does ${a_i}{X_i} \to 0$ almost surely?
Note that the first half is using (the first half of) the Borel-Cantelli lemma and therefore doesn't need the independence; it works as long as the $X_i$ are identically distributed, or even just uniformly bounded in $L^1$.
1d
revised The limit inferior of Borel functions
fix the proof
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asked Is there a name for functions $f$ such that $\{f \le a\}$ is analytic?
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comment The limit inferior of Borel functions
Reposted from mathoverflow.net/q/212709/4832, which was closed but has at least one reopen vote.
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revised The limit inferior of Borel functions
link to proof
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revised The limit inferior of Borel functions
added 25 characters in body
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answered The limit inferior of Borel functions
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comment pointwise almost everywhere convergent subsequence of $\{\sin (nx)\}$
How about you show where you've got so far? meta.math.stackexchange.com/questions/1803/…
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comment Question on Egoroff-like theorem
@AbeerAbassi: When you have got it worked out, you can post your own answer to this question, with your solution.
2d
comment Is 1 + 1 + 1 … a finite number?
This is essentially the reason why mathematical proof usually does not allow for "never-ending processes": proofs must be of finite length. Limiting processes are handled via axioms that specifically allow them, and the Peano axioms don't.
2d
comment Condition implying tightness of sequence of probability measures
Actually, it looks to me like Davide's answer is correct - the given condition is sufficient. And I don't think what you said about a necessary condition is right. Consider a sequence of measures where $\mu_n$ puts mass $n^{-1/2}$ at $n$ and the rest at 0. Is it possible you are thinking of conditions for uniform integrability instead of tightness?
2d
comment Condition implying tightness of sequence of probability measures
I don't know why this was down voted, it looks correct to me.
2d
comment Condition implying tightness of sequence of probability measures
Hmm, so is there a counterexample to the original statement? (By the way, welcome to the site! You can use LaTeX-style formatting, see meta.math.stackexchange.com/questions/5020/…)
2d
comment if the integrals of a non-negative sequence of functions go to zero, does this imply functions go to zero a.e.?
No, the standard counterexample is called the typewriter sequence. This question has certainly been asked many times before, but without a keyboard it's hard for me to search now.
2d
comment If independent r.v. converge in probability to a constant, do they converge almost surely?
A related fact that may interest you: if a series $\sum_n X_n$ of independent random variables $X_n$ converges in probability, then it converges almost surely. See math.stackexchange.com/questions/90116/…
Jul
29
comment Billingsley Exercise 8.8 (Markov Chains)
@Did: No, I don't think you can. Imagine a chain that moves from $i$ to either $0$ or $i+1$, so $p_{i,0} + p_{i,i+1}=1$. If $p_{i,0} \to 0$ sufficiently fast, there may be a positive probability that we never jump to 0. Of course, in that case, the chain drifts off to infinity, so the claim that no state is visited infinitely often is still true.
Jul
29
comment Application of Baire Category theorem
@Gary.: Note that you actually can use the Baire category theorem on the open interval $(a,b)$ - meager / nonmeager are topological properties, so you can use the theorem on any topological space which is homeomorphic to a complete metric space. $(a,b)$ is homeomorphic to $\mathbb{R}$ and with a little work you can even write down a complete metric for $(a,b)$ which induces the usual topology.
Jul
28
comment Is $C[0,1]$ equipped with $\lVert \cdot \rVert_1$ a countable union of nowhere dense sets?
In a more sophisticated sense, the answer has to be "yes" because it certainly isn't Banach, and you didn't use the axiom of choice. If I've done my work correctly, in a model of ZF+DC in which every set of reals has the Baire property, we can show that every incomplete separable normed space is meager. (As a subset of its completion, it has the BP, so it is meager in its completion, and an elementary argument shows it must therefore be meager in itself.)
Jul
28
comment Is $C[0,1]$ equipped with $\lVert \cdot \rVert_1$ a countable union of nowhere dense sets?
@Ian: In some sense they are actually equivalent. You can show that if $X$ is a topological space and $A$ is a dense subset of $X$ then $A$ is meager in itself (in the subspace topology) iff it is meager in $X$.