I am stuck on Problem 4.12 of Karatzas and Shreve's book Stochastic Calculus and Brownian Motion:
Suppose that $\{ \mathbb{P}_n \}$ is a sequence of probability measures on $(C[0, \infty), \mathcal{B} (C[0, \infty)))$ which converges weakly to a probability measure $\mathbb{P}$. Suppose, in addition, that $\{f_n\}$ is a uniformly bounded sequence of real-valued, continuous functions on $C[0, \infty)$ converging to a continuous function $f$, the convergence being uniform on compact subsets of $C[0, \infty)$. Then
$$ \lim_{n \to \infty} \int_{C[0,\infty)} f_n
\, d \mathbb{P_n} = \int_{C[0,\infty)} f
\, d \mathbb{P} .$$
The result is clear if the iterated limit is considered, i.e.
$$ \lim_{m \to \infty} \lim_{n \to \infty} \int_{C[0,\infty)} f_n
\, d \mathbb{P_m} = \int_{C[0,\infty)} f
\, d \mathbb{P} .$$
But I am not sure how we could show this statement. Any ideas?
Best Answer
Fix $\varepsilon$. First use tightness to find a compact subset $K=K(\varepsilon)$ of $C[0,+\infty)$ such that $\mathbb P_n(K)\gt 1-\varepsilon$. Use the uniform convergence of $(f_n)_{n\geqslant 1}$ to $f$ in order to handle the integral of $f_n$ over $K$. Use the uniform bound to handle the integral of $f_n$ over the complement of $K$ (which has a measure $\mathbb P_n$ which does not exceed $\epsilon$).