# Are probability measures on $\mathbb{R}^{d}$ characterized by integrals of $C_{c}$ functions?

In Donald Cohn's Measure Theory, 2nd edition, Lemma 10.3.1 says

if $$\mu$$ and $$\nu$$ are probability measure on $$(\mathbb{R}^{d}, \mathcal{B}(\mathbb{R}^{d}))$$, and $$\int f d\mu = \int f d\nu$$ holds for each bounded continuous function $$f$$ ($$f \in C_{b}$$), then $$\mu = \nu$$.

I was working on a problem about weak convergence (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?) and came up with an idea that we might change $$C_{b}$$ in Lemma 10.3.1 to $$C_{c}$$. Could anyone check my proof?

Suppose $$g \in C_{b}(\mathbb{R}^{d})$$. We want to show: $$\int g d\mu = \int g d\nu$$. We already now $$\int f d\mu = \int f d\nu, \forall f \in C_{c}$$. Now let $$f_{n} \in C_{c}, n = 1, 2, ...$$ Notice the product $$gf_{n} \in C_{c}$$, and so $$\int gf_{n} d\mu = \int gf_{n} d\nu$$. If we can show

$$\int gf_{n} d\mu \xrightarrow{} \int g d\mu$$ and $$\int gf_{n} d\nu \xrightarrow{} \int g d\nu$$,

then the two limits must agree and we are done.

Now let's try to find such $$f_{n}$$. Let $$K$$ be the unit cube in $$\mathbb{R}^{d}$$. Let $$d(x) = \text{inf}\{d(x, y): y \in K\}$$ be the usual distance function to the set $$K$$. Since $$K$$ is closed, we know $$d$$ is continuous and $$d(x) = 0$$ iff $$x \in K$$.

Now let $$K_{n} := \{x: d(x, K) \leq n\}$$, i.e., $$K_{n} = d^{-1}([0, n])$$. Then $$K_{n}$$ is the preimage of a closed set so is closed, and bounded since $$K$$ is bounded.

Now define $$f_{n} := \text{ max}(1 - \frac{1}{n}d(x, K), 0)$$. The maximum function is continuous and so is $$f_{n}$$. Moreover, $$f_{n}$$ takes $$1$$ on $$K$$, takes $$[0, 1)$$ on $$K_{n} - K$$, and $$0$$ on $$K_{n}^{c}$$. As $$n$$ increases, $$f_{n}$$ increases to $$1$$ pointwise, and so $$gf_{n}$$ converges to $$g$$ while being bounded by $$|g|$$. Now the dominated convergence theorem implies $$\int gf_{n} d\mu \xrightarrow{} \int g d\mu$$ and $$\int gf_{n} d\nu \xrightarrow{} \int g d\nu$$.

Thank you for time time!

• Yes, that works. Dec 17, 2022 at 16:59
• @nejimban Thank you!
– Tom
Dec 17, 2022 at 17:23