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Let $X$ be a random variable on $[0,A]$, and $f:[0,A]\to[-B_1,B_2]$ be a continuous function. Let $$g(x) = \frac{g_1(x)}{g_2(x)}$$ where $g_1(x) = E[f(X)\mathbf{1}_{\{f(X)}]$ and $g_2(x) = 1+E[X\mathbf{1}_{\{f(X)\geq 1\}}]$.

My goal is to prove that $g$ is Lipschitz-continous on $[0,E[f(X)]]$.

So far I have, for $\epsilon$ \begin{align} g(x+\epsilon) = \frac{g_1(x) - E[f(X)\mathbf{1}_{\{xX\leq f(X)\leq 1\}}]}{g_2(x) - E[X\mathbf{1}_{\{xX\leq f(X)\leq 1\}}]}. \end{align}

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In general, the function $g$ is not even continuous. E.g., let $A=4$ and $f(x)=x$ for all $x$. Let $X$ be uniformly distributed on $[0,A]=[0,4]$. Then $Ef(X)=2$ and $$g(x)=\frac{1(x\le1)}{1+1(x\le1)},$$ so that $g$ is discontinuous at $1\in[0,2]=[0,Ef(X)]$.

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  • $\begingroup$ Thank you for your answer, however I forgot to explicit the domain of $g$, which is $[0,E[f(X)]]$. Hence in you example $g$ is constant. $\endgroup$
    – Iques
    Commented Apr 29, 2020 at 13:32
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    $\begingroup$ @Iques I think you can take Iosif's example with larger $A$ to make $E[f(x)]$ larger. $\endgroup$
    – Nik Weaver
    Commented Apr 29, 2020 at 14:53
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    $\begingroup$ @Iques : As Nik Weaver noted, even after your change of the question the function $g$ can still be discontinuous, as is now detailed in my answer. More importantly, if your posted question is not what you actually meant, then you (and not the answerer) should take the responsibility for your mistake. That is, you should not change your question so as to invalidate a valid answer. Rather, you may want to post the amended question separately. $\endgroup$
    – Iosif Pinelis
    Commented Apr 29, 2020 at 17:21
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    $\begingroup$ @Iques: to echo that, you are a new contributer and wouldn't know this, but it is considered bad form on Mathoverflow to modify a question after the original version has been answered. Nothing wrong with posting the modified question separately. $\endgroup$
    – Nik Weaver
    Commented Apr 29, 2020 at 21:17

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