Olivier Bégassat
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 Mar 4 awarded Enlightened Mar 4 awarded Nice Answer Feb 20 accepted If $f$ is bounded non-negative $L^1$, is $f\leq g$ a.e. for some continuous integrable $g$? Feb 18 comment If $f$ is bounded non-negative $L^1$, is $f\leq g$ a.e. for some continuous integrable $g$? @A.S. could you add some details on how to construct the transitions while ensuring $g\geq f$? Feb 18 comment If $f$ is bounded non-negative $L^1$, is $f\leq g$ a.e. for some continuous integrable $g$? Wlog $0< f <1$, in which case $E_1=\mathbb{R}$ has infinite measure. I think indeed something like Lusin's theorem may be needed, but I like your idea of approximating by open sets. Feb 18 asked If $f$ is bounded non-negative $L^1$, is $f\leq g$ a.e. for some continuous integrable $g$? Feb 7 awarded Nice Answer Jan 25 awarded Self-Learner Jan 25 awarded Notable Question Jan 24 awarded Nice Answer Jan 18 answered Does free bimodule exist? Dec 16 awarded Popular Question Dec 8 awarded differential-geometry Dec 6 awarded Enlightened Dec 6 awarded Nice Answer Nov 16 comment Prove or Disprove: if $\lim\limits_{n \to \infty} (a_{2n}-a_n)=0,$ then $\lim\limits_{n \to \infty} a_n=0.$ What about the constant sequence $a_n=1$? Nov 8 comment First-order definition of “$f$ is continuous at $x$” using just $<$ @avid19 I see, is that the problem, that there are no such symbols in the language? But how is $\Bbb R$ to be understood in this context? Nov 8 comment First-order definition of “$f$ is continuous at $x$” using just $<$ Wouldn't something like \varphi(x)=\forall\epsilon >0~\exists\delta>0~\forall y~ (x-\delta