Let $\mu_n$ be sequence of probability measures on a polish space $S$ such that for any bounded and continuous $f:S \to \Bbb R$ we have $$\int fd\mu_n \to \int fd\mu$$

Then I have seen in some place claiming the following: $$\int h(x_n,y)\mu_n(dy) \to \int h(x,y)\mu(dy)$$

for $h:\Bbb R \times S \to \Bbb R$ and bounded, continuous with $x_n \to x$.


Edit: Previous answer was bogus.

Let me change your notation a little to let $x_0$ be the limit of the $x_n$.

Yes, this is true. It follows from tightness and the general fact that $h(x_n, \cdot) \to h(x_0, \cdot)$ uniformly on compact sets.

Without loss of generality, let's suppose that $h(x_0, \cdot) = 0$ (replace $h(x,y)$ by $h(x,y) - h(x_0,y)$).

Let $\epsilon > 0$ and let $M$ be the sup norm of $h$. By the easier direction of Prohorov's theorem, the sequence $\{\mu_n\}$ is tight, so there is a compact $K \subset S$ such that for every $n$ we have $\mu_n(K^C) < \epsilon$. Now we have $$\begin{align*}\left|\int h(x_n, y) \mu_n(dy)\right| &\le \int_K |h(x_n,y)|\,\mu_n(dy) + \int_{K^C} |h(x_n, y)| \mu_n(dy) \\ &\le \sup_{y \in K} |h(x_n, y)| + M \epsilon.\end{align*}$$ So it suffices to show that $h(x_n, \cdot) \to 0$ uniformly on $K$. Suppose not; then passing to a subsequence if necessary, we can find $\delta > 0$ so that $\sup_{y \in K} |h(x_n, y)| > \delta$ for all $n$. Thus for each $n$ we can find $y_n \in K$ such that $|h(x_n, y_n)| > \delta$. By compactness, we can pass to a further subsequence so that $y_n$ converges to some $y_0$. Now by continuity of $h$ we have $|h(x_n, y_n)| \to |h(x_0, y_0)| = 0$, a contradiction.

  • $\begingroup$ @podu: Sorry about that. For probability measures it is true. See my edit. $\endgroup$ – Nate Eldredge Jan 19 '15 at 17:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.