# Convergence of a subsequence of a subsequence of distribution functions

I'm trying to find a solution for the following problem:

Let $(\mu_n)_{n\in\mathbb{N}}$ be a sequence of signed (Baire)-measures (of bounded variation) on $[a,b]$ and let $F_{\mu_n}(t):=\mu_n([a,t))$ be the corresponding general distribution functions. By assumption there is a subsequence of a subsequence of $(F_{\mu_n}(t))_{n\in\mathbb{N}}$, denoted (ugly) by $(F_{\mu_{n_{k_l}}}(t))_{l\in\mathbb{N}}$, that converges to a distribution function $F_\mu(t)$ at all points with the exception of points of an at most countable set. Further, by assumption, the sequence $(\mu_n)_{n\in\mathbb{N}}$ does not converge weakly to $\mu$, but I can't imagine this information is even needed here.

Is it now true that for $f\in C^\infty([a,b])$ $$\lim_{l\to\infty}\int_{[a,b]}f'(t)F_{\mu_{n_{k_l}}}(t)\,dt=\int_{[a,b]}f'(t)\lim_{l\to\infty}(F_{\mu_{n_{k_l}}}(t))\,dt= \int_{[a,b]}f'(t)F_\mu(t)\,dt$$ holds?

Should this integral seen as a Riemann-Integral? If so, I'm aware I can exchange the limit and the integral only in the case of uniform continuity, but due to Jegorow the $F_{\mu_{n_{k_l}}}(t)$ converge to $F_\mu(t)$ only uniform almost everywhere if I'm not mistaken.

If the integrals should be seen as Lebesgue-Integrals, then I don't think I can apply dominated convergence or monotone convergence, since I'm dealing only with a subsequence of a subsequence, or am I wrong?

Thank you all very much in advance for your help, this community is great!

Greetings, Florian