# Weak convergence of probability measures and uniform convergence of functions

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?

• Hint 1: use the unfimor convergence of $f_n$ Hint 2: dominated convergence – noctusraid Feb 5 '16 at 17:41
• @noctusraid But uniform convergence only holds for compact subsets. So how can we approximate $C[0, \infty)$ by its compact subsets??? – Richard Feb 5 '16 at 17:44
• Remember that the convergent sequence $\{\Bbb P_n\}$ is necessarily tight. – John Dawkins Feb 5 '16 at 17:57
• @JohnDawkins How can we use the fact that $\{ \mathbb{P}_n \}$ is tight??? – Richard Feb 5 '16 at 18:01

## 1 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 $\epsilon$).

• My main problem now is that I can't bound the integral $\big| \int_K f_n \,d \mathbb{P}_n - \int_K f \, d \mathbb{P} \big|$. We somehow have to relate $\int_K f \,d \mathbb{P}_n$ to $\int_K f \,d \mathbb{P}$. – Richard Feb 6 '16 at 12:40
• Write $\int f_n \,d\Bbb P_n-\int f \,d\Bbb P=\int(f_n-f)d\Bbb P_n+(\int fd\Bbb P_n-\int f\,d\Bbb P)$. The second term on the right converges to $0$ because of the assumed weak convergence. Deal with the first by breaking it up as $\int_K+\int_{K^c}$ and then proceeding as Davide Giraudo suggests. – John Dawkins Feb 6 '16 at 18:10