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Let $X$ be a Polish space and $\mu, \mu_n$ finite signed Borel measures on $X$. Assume that $\mu_n \overset{\ast}{\rightharpoonup} \mu$, i.e., $$ \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu $$ for all bounded continuous functions $f:X \to \mathbb R$. Let $\mu = \mu^+ - \mu^-$ and $\mu_n = \mu_n^+ - \mu_n^-$ be their Jordan decompositions.

Is it true that $\mu^+_n \overset{\ast}{\rightharpoonup} \mu^+$ and $\mu^-_n \overset{\ast}{\rightharpoonup} \mu^-$?

Akira
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  • Parhaps add a link about the type of convergence you are using. – GEdgar Jul 10 '22 at 12:58
  • Hi @GEdgar I work with weak convergence of measures, i.e., $\mu_n \overset{\ast}{\rightharpoonup} \mu$ if and only if $$ \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu $$ for all bounded continuous functions $f:X \to \mathbb R$. – Akira Jul 10 '22 at 16:13
  • I have recently found that it would be true if moreover that $\limsup_n [\mu_n] \le [\mu]$. Here $[\cdot]$ is the total variation norm. – Akira Nov 06 '22 at 09:52

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No, this is not true. Unless $X$ is trivial, you can find a (nonatomic) measure $\lambda$ on $X$ and a sequence $(f_n) \subset L^2(\lambda)$ with $f_n \rightharpoonup 0$ and (say) $f^+_n \rightharpoonup 1$, both in $L^2(\lambda)$. Now, consider $\mu_n = f_n \lambda$.

Finally, if $g_n \rightharpoonup g$ in $L^2(\lambda)$, then $$ \int_X (g_n - g) h \, \mathrm{d}\lambda \to 0 $$ for all $h \in L^2(\lambda)$. Thus, if $\lambda$ is finite, this holds for all bounded, continuous functions $h$. Hence, the measures $g_n \lambda$ converge weak-$*$ towards $g \lambda$.

gerw
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  • It seems I got a contradiction from your construction. Could you elaborate if I misunderstand your ideas? I got that (a) $f_n := f_n^+ - f_n^-$ with $f_n^+, f_n^- \ge 0$, (b) $\int_X f_n^2 \mathrm d \lambda \to 0$ and $\int_X (f_n^+ - 1)^2 \mathrm d \lambda \to 0$, and (c) $\mu_n (B) = \int_B f_n \mathrm d \lambda$ for all $B \in \mathcal B(X)$. Then (b) implies $\int_X (f^+_n)^2 \mathrm d \lambda \to 0$ and $\int_X (f^-_n)^2 \mathrm d \lambda \to 0$. This implies $f_n^+ \rightharpoonup 0$ in $L^2(\lambda)$, which is a contradiction. – Akira Jul 09 '22 at 17:08
  • No, "$\rightharpoonup$" indicates weak convergence. – gerw Jul 09 '22 at 17:50
  • I don't get how weak convergence of $f_n, f_n^+$ comes into the play. Could you expand your answer a little bit? – Akira Jul 09 '22 at 23:18
  • I have added some further explanation. – gerw Jul 10 '22 at 12:20
  • Could you confirm if my understanding is correct? (1.) $g_n \lambda$ is a measure on $X$ defined by $g_n \lambda (B) := \int_B g_n \mathrm d \lambda$ for every Borel set $B$. This means $g_n$ is the Radon-Nikodym derivative of $g_n \lambda$ w.r.t. $\lambda$. (2.) $\rightharpoonup$ is the convergence in the weak topology $\sigma(L_2 (\lambda), L_2 (\lambda)^)$. (3.) $L_2 (\lambda)$ is isometrically isomorphic to $L_2 (\lambda)^$. So for each $\varphi \in L_2 (\lambda)^*$, there a unique $h \in L_2 (\lambda)$ such that $\varphi (f) = \int_X hf \mathrm d \lambda$ for all $f \in L_2 (\lambda)$. – Akira Nov 03 '22 at 16:27
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    Yes, this seems to be correct. – gerw Nov 03 '22 at 18:12
  • Thank you so much! It's until now that I can understand what you meant... – Akira Nov 03 '22 at 18:13