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!

  • 2
    $\begingroup$ Yes, that works. $\endgroup$
    – nejimban
    Commented Dec 17, 2022 at 16:59
  • $\begingroup$ @nejimban Thank you! $\endgroup$
    – Tom
    Commented Dec 17, 2022 at 17:23


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