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Aug
6
comment How to `bound' $L^\infty$ by the constant function $1$
But this bound only works almost everywhere. Sure, you can find a representative of $f$ from same equivalence class for which this works, but in general $f$ can be unbounded. Right?
Aug
1
awarded  Benefactor
Aug
1
accepted Disintegration theorem, a reference needed
Aug
1
comment Disintegration theorem, a reference needed
@ByronSchmuland: Thanks, I will also look that up :-)
Aug
1
comment Disintegration theorem, a reference needed
@ByronSchmuland: I'm looking for something covering the existence of disintegrations in atleast as general setting as presented in Wikipedia, which is Radon spaces. Particularly I would be prefer to avoid probabilistic conditioning approaches and keep it on a general level. I think Fremlin's book was a perfect match for this search, but in any case, I'm grateful for every answer made in this topic.
Aug
1
comment Disintegration theorem, a reference needed
So far Fremlin's book has been the most impressive and it is exactly what I was looking for. Not only is this opus amazing but so are his other books of the same 'series'. I also looked up rest of your suggestions and they were useful too, thank you. The bounty can be awarded after 1 hour, so until then if nothing better (which I doubt) appears then it surely belongs to this answer.
Jul
31
comment Disintegration theorem, a reference needed
@EdGorcenski. You ignored the second part of the sentence which gave some specification on its style. The book in question is (most likely, judging from its appearance) written with a typewriter machine.
Jul
31
comment Disintegration theorem, a reference needed
Since none of the (so far) provided books have met my interests and this is a rather important matter for me, I have raised a 50 rep bounty for this question.
Jul
31
awarded  Promoter
Jul
31
comment The metrizable space may be not locally compact
@Paul. Since in metric spaces compactness is equivalent with sequential compactness then it suffices. Take any open ball, it contains Cauchy sequences that have a limit in $\mathbb{R}$ but not in $\mathbb{Q}$ (a sequence that converges to an irrational number), then by closuring this open ball in $\mathbb{Q}$ you at most obtain more rationals in the set and not these limits a priori. Thus the closure of every open ball has sequences with no convergent subsequences and is hence non-compact.
Jul
31
comment The metrizable space may be not locally compact
You have sequences that lie entirely in $\mathbb{Q}$ but converge to irrationals in $\mathbb{R}$. By taking a closure of a set in the relative topology of $\mathbb{Q}$ you at most gain more rationals to the set, not these limits for example. For this reason, from the closure of every open ball in $\mathbb{Q}$ you find sequences with no convergent subsequences.
Jul
31
comment Disintegration theorem, a reference needed
@did: Yes, I have that book in my shelf and I'm reading it as we speak. However, I couldn't find anything related to what you suggest. My best guess would be that this topic could be covered under Chapters 6.33-34, but it looks like they aren't. Maybe he covers these topics with different terminology, but I highly doubt it would be dealt before the Radon-Nikodym section. Altogether, so far I haven't been successful in finding this book useful in this particular context.
Jul
30
comment In mathematics, what is meant by induction?
The Wikipedia article on this topic can be useful to read: en.wikipedia.org/wiki/Mathematical_induction
Jul
28
comment Disintegration theorem, a reference needed
@StefanHansen: Could be, I'm not that familiar with conditional probability theory. If you can point some sources then I would be glad to go them through. And for others: I'm also looking for sources that deal with general metric spaces (locally compact Polish, compact?) with the push-forward framework and disintegration. Mainly in non probabilistic setting.
Jul
27
asked Disintegration theorem, a reference needed
Jul
23
comment If a subset of $\mathbb{R}$ is closed and bounded with respect to a metric equivalent to the Euclidean metric, must it be compact?
Just an idea: what if $\hat{d}$ would be such that every bounded set would also be totally bounded respect to this metric (as in the Euclidean case). Would this be sufficient for these statements to hold? And would it also be a necessary condition?
Jul
22
comment Hausdorff Measure
By line do you mean a subset of $\mathbb{R}^{2}$ or an interval i.e. a subset of $\mathbb{R}$? The answer is different depending on what you mean.
Jul
12
comment Proving a theorem from topology
To me this looks like a definition rather than a theorem. Also, where is the proof that you can not follow?
Jul
4
comment How to solve infinite repeating exponents
@Matt. Dito. And even further, you may notice that $e^{\frac{ln(2)}{2}}=e^{ln(\sqrt{2})}=\sqrt{2}$.
Jul
4
comment Not every metric is induced from a norm
Did you mean instead that any metric not satisfying either of those can not come from a norm? And not the other way around? And also, these two conditions on the metric are sufficient to define a norm..?