1

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 $x_n \to x$.

It looks like convergence in product topology is used, I am not much aware of such things

Anonymous
  • 2,039
  • 2
  • 16
  • 22
  • 1
    Can you prove it for indicator/simple functions, and then extend to bounded continuous functions? – Alex R. Jan 19 '15 at 06:52

1 Answers1

1

By Prokhorov's Theorem (http://en.wikipedia.org/wiki/Prokhorov%27s_theorem), the sequence $(\mu_n)_n$ is (uniformly) tight, i.e. for every $\varepsilon > 0$, there is a compact set $K_\varepsilon \subset S$ such that $\mu_n ( S\setminus K_\varepsilon) < \varepsilon$ for all $n \in \Bbb{N}$.

Now use that the set $L := \{x_n \mid n \in \Bbb{N}\} \cup \{x\}$ is also compact, so that $L \times K_\varepsilon$ is compact.

We can now use that continuous functions on compact sets are uniformly continuous.

Together with the boundedness of $h$, which ensures that

$$ \bigg| \int_{S \setminus K_\varepsilon} h(x_n, y) \, d\mu_n(y)\bigg| \leq C \cdot \mu_n (S \setminus K_\varepsilon) \leq C \cdot \varepsilon, $$ this should allow you to conclude the proof.

EDIT: Putting all pieces together, we arrive at

\begin{eqnarray*} & & \left|\int h\left(x_{n},y\right)\,{\rm d}\mu_{n}\left(y\right)-\int h\left(x,y\right)\,{\rm d}\mu\left(y\right)\right|\\ & \leq & \left|\int_{S\setminus K_{\varepsilon}}h\left(x_{n},y\right)\,{\rm d}\mu_{n}\left(y\right)\right|+\left|\int_{K_{\varepsilon}}h\left(x_{n},y\right)-h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)\right|+\left|\int_{K_{\varepsilon}}h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)-\int_{S}h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)\right|+\left|\int_{S}h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)-\int_{S}h\left(x,y\right)\,{\rm d}\mu\left(y\right)\right|\\ & \leq & C\cdot\varepsilon+\sup_{y\in K_{\varepsilon}}\left|h\left(x_{n},y\right)-h\left(x,y\right)\right|+\left|\int_{S\setminus K_{\varepsilon}}h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)\right|+\left|\int_{S}h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)-\int_{S}h\left(x,y\right)\,{\rm d}\mu\left(y\right)\right|\\ & \leq & 2C\cdot\varepsilon+\sup_{y\in K_{\varepsilon}}\left|h\left(x_{n},y\right)-h\left(x,y\right)\right|+\left|\int_{S}h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)-\int_{S}h\left(x,y\right)\,{\rm d}\mu\left(y\right)\right|\\ & \xrightarrow[n\to\infty]{} & 2C\cdot\varepsilon, \end{eqnarray*} where the last step used the uniform continuity of $h$ on $L\times K_\varepsilon$ and the assumption on the convergence $\mu_n \to \mu$ (in the sense of your assumption). As $\varepsilon > 0$ was arbitrary, this establishes the claim.

PhoemueX
  • 36,211