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I'm trying to prove the first part of Proposition 8.1.8 in V.I.Bogachev, Measure Theory 2:

A sequence of signed measures $\mu_n$ on the interval $[a,b]$ converges weakly to a measure $\mu$ precisely when $\sup_n\lvert|\mu_n|\rvert<\infty$ and every subsequence in the sequence of the distribution functions $F_{\mu_n}$ of the measures $\mu_n$ contains a further subsequence convergent to $F_\mu$ at all points, with the exception of points of an at most countable set.

I'm struggling with the direction $\Leftarrow$ and already asked a question about this here, but I'm not sure if I did mention everything needed so let me cite the first few lines of the proof:

Suppose that the measures $\mu_n$ are uniformly bounded and satisfy the indicated condition with subsequences, but do not converge weakly to $\mu$. Since every continuous function $f$ can be uniformly approximated by smooth functions, we obtain, taking into account the boundedness of $\lvert|\mu_n|\rvert$, that there exists a smooth function $f$ such that the integrals of $f$ against the measures $\mu_n$ do not converge to the integral of $f$ against $\mu$. Passing to a subsequence, we may assume that the difference between the indicated integrals remains greater than some $\delta>0$. Passing to a subsequence once again we can assume that $\lim_{n\to\infty}F_{\mu_n}=F_\mu$ everywhere, with the exception of finiteley or countably many points.

No problems for me so far. Please notice $F_{\mu_n}(t):=\mu_n([a,t))$.

The functions $F_\mu$ and $F_{\mu_n}$ are constant on $(b,+\infty)$, hence $\mu([a,b])=\lim_{n\to\infty}\mu_n([a,b])$. Then the integration by parts formula yields that the right-hand side of the equality $$\int_a^bf(t)\mu_n(dt)=f(b)F_{\mu_n}(b+)-\int_a^bf'(t)F_{\mu_n}(t)dt$$ converges to $$ f(b)F_{\mu}(b+)-\int_a^bf'(t)F_{\mu}(t)dt=\int_a^bf(t)\mu(dt)$$ which leads to a contradiction.

My problem here is I don't see why the last integral in the first line converges to the first integral in the second line. I don't see any possibility to apply monotone or dominated convergence, so I have no idea what to do.

Any help would be highly appreciated, I'm really stuck here. Thank you very much in advance.

Edit: If this is trivial, please just give me a short hint. This is driving me crazy!

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    $\begingroup$ If $\vert F_{\mu_n}\vert \leq g$ a. e. with $g\in\mathcal{L}^1$, you can apply the dominated convergence theorem to pass the limit into the integral and have $\lim_{n\rightarrow \infty}\int_a^b f'(t)F_{\mu_n}(t)dt=\int_a^b f'(t) \lim_{n\rightarrow \infty}F_{\mu_n}(t)dt=\int_a^bf'(t)F_{\mu}(t)dt$. $\endgroup$ – Luc M May 30 '15 at 18:15
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Ok, I got this myself I think. For those who are interested in this problem, here is my solution:

First, notice that the little imprecise wording "Passing to a subsequence, we may assume that the difference between the indicated integrals remains greater than some δ>0" means

$$\exists\,(\mu_{n_k})_{k\in\mathbb{N}},\,\exists\,\varepsilon_0>0, \,\exists\,f\in\mathcal{C}^\infty(X),\,\exists\,K\in\mathbb{N}\,:\,\forall\,k\geq K\,:\,\\ \bigg|\int_Xf\,d\mu_{n_k}-\int_Xf\,d\mu\bigg|\geq\varepsilon_0$$

This is to get the desired contradiction. Further, since $\sup_{n\in\mathbb{N}}\lvert|\mu_n|\rvert=:C<\infty$, one can choose $g(x):=C$ and apply dominated convergence on $|F_{\mu_{n_{k_l}}}(t)|\leq C$ to obtain

$$ \lim_{l\to\infty}\int_a^bf'(t)F_{\mu_{n_{k_l}}}(t)\,dt=\int_a^bf'(t)F_\mu(t)\,dt.$$

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