# 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]

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?

• @OliverDíaz Thanks a lot!
– Tom
Dec 16, 2022 at 21:08

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 $$\mu_n \stackrel{\mathrm{w}}{\rightarrow} \mu$$.

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?