Weak convergence is defined in terms of $C_{b}$ functions, but for $\mathbb{R}^{d}$ why is it sufficient to show convergence for $C_{c}$ functions? [duplicate]

Question

Let $(X, \mathcal{F})$ be a measurable space, and $\mu_{n}, \mu$ probability measures on it. $\mu_{n}$ is said to converge weakly to $\mu$ if for any bounded continuous functions $f$ on $X$, $\int f d\mu_{n} \xrightarrow{} \int f d\mu$.

The professor mentioned if $X = \mathbb{R}^{d}$ and $\mathcal{F}$ is the Borel sigma algebra, then it is enough to check the convergence of integrals for any compactly supported continuous function $f$. But why is this true?

Answer 1

This is because all probability measures on $\mathbf{R}^d$ are inner-regular.

The relevant consequence of this fact is that for all $\epsilon > 0$, there exists a compact $K \subseteq \mathbf{R}^d$ such that $\mu(K^c) < \epsilon$.

Now, since $\mu_n(K) \rightarrow \mu(K)$ (there's a small argument here, since $\mathbf{1}_K$ is not continuous, but it's not difficult), we have that $$ \limsup_{n\rightarrow \infty}\mu_n(K^c) = 1 - \liminf_{n\rightarrow\infty}\mu_n(K) = 1 - \mu(K) = \mu(K^c) < \epsilon. $$

Therefore, for any $f \in C_b(\mathbf{R}^d)$, we have that \begin{align*} \limsup_{n\rightarrow\infty}|\mu(f) - \mu_n(f)| &\leq \limsup_{n\rightarrow\infty}|\mu(f\mathbf{1}_K) - \mu_n(f\mathbf{1}_K)| + \limsup_{n\rightarrow\infty}\lVert{f} \rVert_\infty (\mu(K^c) + \mu_n(K^c)) \\ &\leq 2\epsilon \lVert{f} \rVert_\infty. \end{align*}

(Again, there's a detail here where $f \mathbf{1}_K$ is not continuous and so not in $C_c$, but this is not hard to fix - the important thing is that it is compactly supported)

Since this holds for all $\epsilon > 0$, we have that $\mu_n(f) \rightarrow \mu(f)$ and so $\mu_n \stackrel{\mathrm{w}}{\rightarrow} \mu$.

Answer 2

After discussing the problem with others I'm now trying to show the progress I make by far.

Suppose $g \in C_{b}(\mathbb{R}^{d})$. We want to show

(1): $\int g d\mu_{n} \xrightarrow{} \int g d\mu$.

Now fix any $\epsilon > 0$. Since there is a sequence of closed cubes increasing to the whole space, by lower continuity of measure there is a closed cube $K$ such that

(2): $\mu(K) \geq 1-\epsilon$.

Now fix $\delta$ to be a small positive number. Consider the set $O = \{x: d(x, K) \leq \delta\}$, where $d$ is the usual distance function, continuous because $K$ is closed (Distance to a closed set is continuous.). It's also an easy result that $d(x, K) = 0$ iff $x \in K$. Since it's continuous and $[0, \delta]$ is closed, we know $O$ is a closed set. Since $K$ is bounded, $O$ is bounded. And so $O$ is a compact set.

Now define a function $f = \text{ max}(1 - \frac{1}{\delta}d(x, K), 0)$. The maximum function is continuous and so $f$ is continuous. Furthermore, on $K$, $f = 1$; on $O - K$, $f \in [0, 1)$; on $O^{c}$, $f = 0$.

Notice $f \in C_{c}$ and $f$ is no larger than the indicator function of $O$ so we have $\mu_{n}(O) \geq \int f d\mu_{n}, \forall n$. By our assumption, $\int f d\mu_{n} \xrightarrow{} \int f d\mu \geq \int_{K} f d\mu = \mu(K) \geq 1-\epsilon$. So there is an integer $N$ such that for all $n \geq N$, $\mu_{n}(O) \geq \int f d\mu_{n} \geq 1 - 2\epsilon$. This means $\mu_{n}(O^{c}) \leq 2\epsilon, n \geq N$.

Now up to discarding $\mu_{1}, ..., \mu_{N-1}$ the sequence of measures $\mu_{n}$ is uniformly tight. What next?