# Thread: Weak convergence of measures

1. ## Weak convergence of measures

Let $(X,d)$ a metric space, $\mathcal{B}(X)$ the algebra of its Borel sets, $(\mathbb{P}_n)_n$ a sequence of probability measures on $(X,\mathcal{B}(X))$ and $C_b(X)$ the space of bounded continuous functions on $X$.
We say the sequence $\mathbb{P}_n$ converges weakly (e.g. in distribution) to the probability measure $\mathbb{P}$ if, for each $f\in C_b(X)$:
$\lim_{n\rightarrow\infty}\int_Xf(x)d\mathbb{P}_n(x )=\int_Xf(x)d\mathbb{P}(x)$

Now suppose that, given a sequence of measures $(\mathbb{P}_n)_n$ on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, we have:
$\lim_{n\rightarrow\infty}\int_{\mathbb{R}}f(x)d\ma thbb{P}_n(x)=\int_{\mathbb{R}}f(x)d\mathbb{P}(x)$ for every infinitely differentiable function $f$ with compact support.
How to you prove this implies that $(\mathbb{P}_n)_n$ converges weakly to $\mathbb{P}$?

The Weierstrass theorem only applies to continuous functions on a compact, so it can't be used here. I wouldn't be surprised if this is a classic, but I would greatly appreciate some help.

2. Originally Posted by akbar
The Weierstrass theorem only applies to continuous functions on a compact, so it can't be used here.
Yes it can. Let $f\in\mathcal{C}_b(\mathcal{R})$.

Let $\varepsilon>0$.

There exists a segment $K=[a,b]$ such that $\mu(K^c)\leq\varepsilon$. Choose $\theta$ to be a $[0,1]$-valued $\mathcal{C}^\infty$ function that equals 1 on $K$ and 0 on $(K')^c$ where $K'=K+[-1,1]=[a-1,b+1]$ (for instance). Then, since $\theta\in\mathcal{C}^\infty_c$, using the assumption there exists $n_0$ such that, for $n\geq n_0$, $\int (1-\theta) d\mu_n\leq \varepsilon+\int (1-\theta) d\mu\leq \varepsilon+\mu(K^c)\leq 2\varepsilon$.

Now write $f=f\theta+f(1-\theta)$. We have $\left|\int f(1-\theta) d\mu_n\right|\leq 2 \|f\|_\infty \varepsilon$. Same with respect to $\mu$, hence $\left|\int f(1-\theta)d\mu_n-\int f(1-\theta)d\mu\right|\leq 4\|f\|_\infty \varepsilon$.
And for the first term, Stone-Weierstrass provides a polynomial $P$ such that $|f(x)-P(x)|\leq \varepsilon$ for $x\in K'$, hence $|f(x)\theta(x)-P(x)\theta(x)|\leq \varepsilon$, and $\psi=P\theta$ is in $\mathcal{C}^\infty_c$. Hence $\left|\int f\theta d\mu_n - \int \psi d\mu_n\right|\leq \varepsilon$, and we also have $\left|\int \psi d\mu_n - \int \psi d\mu\right|<\varepsilon$ for $n\geq n_1(\geq n_0)$ using the assumption. Finally, $\left|\int \psi d\mu -\int f\theta d\mu\right|\leq \|\psi-f\theta\|_\infty\leq \varepsilon$ like with $\mu_n$.

Glue together the above pieces, and you get what you want for $n\geq n_1$: $\left|\int f d\mu_n-\int f d\mu\right|\leq C\varepsilon$

3. Finally made some sense out of it. The trick with $f=f\theta+f(1-\theta)$ was hard to improvise...
Thanks again.