Convergence of integrals under weak convergence of measure and compact convergence

functional-analysisweak-convergence

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

Thanks in advance 🙂

Best Answer

Finally got it:

$|\int_\Omega (f_n-f)dP_n|\leq \int_{K_\epsilon} |f_n-f|dP_n +\int_{K^c_\epsilon} |f_n-f|dP_n \leq \sup_{\omega\in K_\epsilon}|f_n(\omega)-f(\omega)|+\epsilon||f_n-f||_\infty.$

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 $f$.

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