# Convergence of integrals under weak convergence of measure and compact convergence

I'm trying to solve Problem 2.4.12 on page 64 of Karatzas-Shreve's book "Brownian motion and stochastic calculus":

My attempt is to use triangle inequality (denoting $\Omega=C[0,\infty)$) $$|\int_\Omega f_ndP_n-\int_\Omega fdP|\leq |\int_\Omega (f_n-f)dP_n|+|\int_\Omega fdP_n-\int_\Omega fdP|\quad\star,$$ and estimate the first term with dominated convergence, the second using weak convergence of the measures.

For $\epsilon>0$, since $(P_n)_n$ is tight (by Prohorov thm) I can choose a compact $K\subset \Omega$ such that $P_n(K)\geq 1-\epsilon$. Fix $m\geq1$, then by dominated convergence for $n$ large enough I have $$|\int_\Omega (f_n-f)dP_m|=|\int_K (f_n-f)dP_m|+|\int_{K^c} (f_n-f)dP_m| \leq \epsilon+\epsilon\sup_{n\geq1,\omega\in K^c}|f_n(\omega)-f(\omega)|.$$

1. To conclude I need $f$ to be bounded. Does this follow from the fact that $f$ is the compact-limit of uniformly bounded functions? (Note that I need boundedness also to use weak convergence on the second term of $\star$)

2. Is it correct that if $\forall m\geq1$ $|\int_\Omega (f_n-f)dP_m|\to_{n\to\infty}0$, then $|\int_\Omega (f_n-f)dP_n|\to_{n\to\infty}0?$ To be fair I think it's true only if the first convergence is uniform in $m$.

• Re your 2nd question: What's wrong about doing the estimate for $m=n$? – saz Jul 29 '18 at 5:39
• @saz: by fixing $m\geq1,\epsilon>0$ I found $N(\epsilon,m)\geq1$ such that $\forall n\geq N(\epsilon,m),$ $\int_\Omega(f_n-f)dP_m<\epsilon$. My problem is that $N(\epsilon,m)$ depends on $m$. How do I know that $\sup_m N(\epsilon,m)$ is finite? – Demetrio Masciurett Jul 29 '18 at 8:58
• @Bob: thanks, since I have point-wise convergence and boundedness I even don't need to separate the integral in $K$ and $K^c$: I can apply dom convergence directly on $\Omega$, right? – Demetrio Masciurett Jul 29 '18 at 9:45
$|\int_\Omega (f_n-f)dP_n|\leq \int_{K_\epsilon} |f_n-f|dP_n +\int_{K^c_\epsilon} |f_n-f|dP_n \leq \sup_{\omega\in K_\epsilon}|f_n(\omega)-f(\omega)|+\epsilon||f_n-f||_\infty.$
Since $\epsilon$ is arbitrary, letting $n\to \infty$ the last term goes to zero by uniform compact convergence of $(f_n)_{n\geq 1}$ to $f$.