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Jan
20
accepted If $\int f d\mu = 1$ with $\mu$ probability measure, then $f(x)=1$ for $f\mu$-a.e. $x$?
Jan
20
comment If $\int f d\mu = 1$ with $\mu$ probability measure, then $f(x)=1$ for $f\mu$-a.e. $x$?
Thanks, I am always stupid, that was easy.
Jan
20
asked If $\int f d\mu = 1$ with $\mu$ probability measure, then $f(x)=1$ for $f\mu$-a.e. $x$?
Nov
23
accepted Does equality of push-forward of measures imply equality of measures?
Nov
23
comment Does equality of push-forward of measures imply equality of measures?
Yes, you got it, I am very convinced now. Thank you very much.
Nov
23
comment Does equality of push-forward of measures imply equality of measures?
Nice reply, thanks. So actually I should also assume that $x \mapsto F(\cdot, x)$ is non-constant: this would rule out your counterexample, right? Which would be the answer in that case? Thank you again for your help.
Nov
23
asked Does equality of push-forward of measures imply equality of measures?
Nov
10
comment Unbounded function of bounded variation (in $\mathbb R^d$, $d>1$)
You are right, I love it :-)! Unfortunately, I am not used to this kind of reasoning (I mean using scales and "units of measurement") but they are definitely very useful and I should learn how to use them. Beautiful trick, thank you again.
Nov
10
comment Unbounded function of bounded variation (in $\mathbb R^d$, $d>1$)
Thank you, Giovanni, this is also nice and easy to take in mind: you are right, I should have thought to $W^{1,1}$. By the way, I hope you won't get angry, if I accept David's answer.
Nov
10
accepted Unbounded function of bounded variation (in $\mathbb R^d$, $d>1$)
Nov
10
comment Unbounded function of bounded variation (in $\mathbb R^d$, $d>1$)
Beautiful answer, very useful. Thanks a lot, I think that from now on I have no more excuses: I cannot forget your answer and I will have no longer doubts on this question. Thanks again.
Nov
10
asked Unbounded function of bounded variation (in $\mathbb R^d$, $d>1$)
Oct
9
awarded  Popular Question
Oct
9
accepted Assume $\sup_n \int_\Omega f_n \, d\mu < + \infty$. Does it follow $\sup_n f_n(x) < +\infty$ a.e.?
Oct
9
comment Assume $\sup_n \int_\Omega f_n \, d\mu < + \infty$. Does it follow $\sup_n f_n(x) < +\infty$ a.e.?
Thanks, you are absolutely right. It was easy... The piecewise constant functions we construct have the property that $\sup_n f_n(x)=+\infty$ for every $x\in [0,1]$. Thanks a lot and sorry for the easy question.
Oct
9
asked Assume $\sup_n \int_\Omega f_n \, d\mu < + \infty$. Does it follow $\sup_n f_n(x) < +\infty$ a.e.?
Jul
13
comment Rigor in Banach contraction principle
Sure, (Lipschitz) continuity is given by contraction property (this is what I used in the last line).
Jul
13
comment Rigor in Banach contraction principle
Done, hope it is more clear now.
Jul
13
revised Rigor in Banach contraction principle
added 196 characters in body
Jul
13
answered Rigor in Banach contraction principle