Romeo
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 Apr 10 awarded Yearling Mar 15 awarded Popular Question 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.