# Weak convergence of sequence of measures iff every subsequence in sequence of distribution functions contains an a.e. convergent subsequence

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!

• 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 $\lim_{n\rightarrow \infty}\int_a^b f'(t)F_{\mu_n}(t)dt=\int_a^b f'(t) \lim_{n\rightarrow \infty}F_{\mu_n}(t)dt=\int_a^bf'(t)F_{\mu}(t)dt$. – Luc M May 30 '15 at 18:15

## 1 Answer

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

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