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

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 :)

  • $\begingroup$ You know that the sequence is uniformly bounded, so the same bound holds also for the pointwise limit $\endgroup$ – Bob Jul 28 '18 at 21:23
  • $\begingroup$ Re your 2nd question: What's wrong about doing the estimate for $m=n$? $\endgroup$ – saz Jul 29 '18 at 5:39
  • $\begingroup$ @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? $\endgroup$ – Demetrio Masciurett Jul 29 '18 at 8:58
  • $\begingroup$ @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? $\endgroup$ – Demetrio Masciurett Jul 29 '18 at 9:45

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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