Reputation
2,802
Top tag
Next privilege 3,000 Rep.
Cast close & reopen votes
Badges
2 7 30
Newest
 Yearling
Impact
~57k people reached

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.