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I'm struggling in finding the hypotheses for weak convergence to imply setwise convergence, for a sequence of induced measures.

Consider a standard probability triple $(\Omega,\mathcal{B},\mu)$. Let $f_n\rightarrow f$ pointwise. Let $\mu_f$ be the measure induced by f and $\mu_n$ be the measure induced by $f_n$ for each n. Finally let $\mu_n$ converge weakly to $\mu_f$, that is, $\int_{\Omega} h \,d\mu_n\rightarrow \int_\Omega h \, d\mu_f$ for all bounded continuous $h$.

Under what additional hypothesis on $f$ and $f_n$, $\mu_n$ converges setwise to $\mu_f$: $\lim_n \mu_n(A)=\mu_f(A)$ for all $A\in\mathcal{B}$?

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    $\begingroup$ Your notation is irritating. You use $f$ for different types of functions, f.i. in $\int fd\mu_f$. $\endgroup$
    – Dieter Kadelka
    Commented Mar 25, 2022 at 16:42
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    $\begingroup$ Generally, $\int g~\mathrm d\mu_f=\int g\circ f~\mathrm d\mu$, so pointwise convergence implies weak convergence of the distributions by the dominated convergence theorem. So the question boils down to conditions for pointwise convergence to imply setwise convergence of distributions. Setwise convergence of distributions seems to be too strong a requirement to be satisfied in any nontrivial situation. $\endgroup$
    – Michael Greinecker
    Commented Mar 25, 2022 at 17:01
  • $\begingroup$ I apologize for the notation, thanks for the corrections. $\endgroup$
    – Gioppa
    Commented Mar 28, 2022 at 15:28
  • $\begingroup$ @MichaelGreinecker I was looking for conditions on the functions $f_n$ and $f$. Maybe smoothness of something similar. $\endgroup$
    – Gioppa
    Commented Mar 28, 2022 at 15:31
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    $\begingroup$ @Gioppa Even being constant would not be enough. The sequence of functions with constant value $1/n$ converges uniformly, but the distributions do not converge setwise. $\endgroup$
    – Michael Greinecker
    Commented Mar 28, 2022 at 16:28

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