Borel Measure, Support and Regularity Property

Definition 1. If $(X,\tau)$ is a topological space, then the Borel $\sigma$-field $\mathcal{B}$ of $X$ is the $\sigma$-field generated by the open sets of $X$ and we say a measure $\mu$ defined on $(X,\mathcal{B})$ is a Borel measure.

Proposition 1. Suppose that $\Sigma$ is $\sigma$-field of subsets of a metrizable space $(X,d)$. The following are equivalent.

(i) $\Sigma=\mathcal{B}$, the Borel $\sigma$-field of $X$.

(ii) $\Sigma=\Sigma_1$, the smallest $\sigma$-field for which every continuous real-valued function on $X$ is measurable.

(iii) $\Sigma=\Sigma_2$, the smallest $\sigma$-field for which every lower semi-continuous real-valued function on $X$ is measurable.

Proof: Let $C$ be the collection of continuous real-valued function on $X$, $LSC$ be the lower semi-continuous real-valued function on $X$ and $B$ be the Borel measurable real-valued function on $X$. Clearly, $C\subseteq LSC$. If $f$ is lower semi-continuous, for each $c\in\mathbb{R}$, since every sublevel set of a lower semi-continuous function is closed, namely, $\{f\le c\}$ is closed and then Borel measurable. Hence $LSC\subseteq B$. Since

$$
\Sigma_1 = \sigma\left(\bigcup_{f\in C} \{f^{-1}(A) ; A\in \mathcal{B}\}\right), \Sigma_2 = \sigma\left(\bigcup_{f\in LSC} \{f^{-1}(A) ; A\in \mathcal{B}\}\right), \mathcal{B} = \sigma\left(\bigcup_{f\in B} \{f^{-1}(A) ; A\in \mathcal{B}\}\right),
$$

we have $\Sigma_1\subseteq \Sigma_2\subseteq \mathcal{B}$. Finally, if $A$ is closed, then
$$
A=\{x\in X:d(x,A)=0\},
$$
so that $A\in\Sigma_1$, and $\mathcal{B}\subseteq\Sigma_1$. $\square$

Support of Borel Measure

If $\mu$ is a Borel measure on a separable metrizable space $(X,\tau)$, we can define its support $\operatorname{supp}(\mu)$.

Definition 2. If $\mu$ is a finite Borel measure on a topological space $(X,\tau)$, a closed subset $C$ of $X$ is the support of $\mu$, if $\mu(X\setminus C)=0$ and $C$ is the smallest closed subset of $X$ with this property.

Proposition 2. Suppose that $\mu$ is a Borel measure on a separable metrizable space $(X,\tau)$. Then $\mu$ has a support.

Proof: Since separable metrizable space is second countable, $(X,\tau)$ is second countable. Let $(U_n)_{n=1}^{\infty}$ be a base of open sets, and let

$$
K=\{n\in\mathbb{N};\mu(U_n)=0\}.
$$
Let

$$
U=\bigcup_{n\in K} U_n.
$$
Then $\mu(U)=0$, and we need to show that $U$ is the largest open subset of $X$ with this property.

If there is a set $V$ such that $\mu(V)=0$ but $V\not\subset U$. There is a set $T\subseteq \mathbb{N}$ such that

$$
V=\bigcup_{n\in T} U_n,
$$

and there exists $n_0\in \mathbb{N}$ such that $n_0\in T$ and $n_0\notin K$. Then

$$
\mu(V)\ge \mu(U_{n_0})>0,
$$

which is a contradiction. Hence $U$ is the largest open subset of $X$ such that $\mu(U)=0$. Thus $X\setminus U$ is the support of $\mu$. $\square$

Proposition 3. Suppose that $\mu$ is a Borel measure on a metrizable space $(X,\tau)$ with support set nonempty and that $x\in X$. Then $x\in \operatorname{supp}(\mu)$ if and only if $\mu(N)>0$ for each open neighborhood of $x$.

Proof: We first prove if $x \in \operatorname{supp}(\mu)$, then every open neighborhood of $x$ has positive measure.

Assume for contradiction that there exist $x \in \operatorname{supp}(\mu)$ and an open neighborhood $N$ of $x$ such that
$$
\mu(N)=0.
$$

Since $N$ is open and $x \in N$, let
$$
F := \operatorname{supp}(\mu)\setminus N.
$$

Then $F$ is closed, because $\operatorname{supp}(\mu)$ is closed and $N$ is open. Moreover,
$$
X\setminus F
= X\setminus (\operatorname{supp}(\mu)\setminus N)
= (X\setminus \operatorname{supp}(\mu)) \cup N.
$$

Hence,
$$
\mu(X\setminus F)
\le \mu(X\setminus \operatorname{supp}(\mu)) + \mu(N)
= 0 + 0
= 0.
$$

So $F$ is also a closed set whose complement has measure zero. But $x \notin F$, and therefore

$$
F \subsetneq \operatorname{supp}(\mu),
$$
which contradicts the minimality of $\operatorname{supp}(\mu)$.

Next we prove if every open neighborhood of $x$ has positive measure, then $x \in \operatorname{supp}(\mu)$.

Now assume that for every open neighborhood $N$ of $x$, one has $\mu(N)>0$. Assume for contradiction that $x \notin \operatorname{supp}(\mu)$. Since $\operatorname{supp}(\mu)$ is closed,
$$
N := X\setminus \operatorname{supp}(\mu)
$$
is open, and $x \in N$. Hence $N$ is an open neighborhood of $x$. But by the definition of the support,

$$
\mu(X\setminus \operatorname{supp}(\mu))=0,
$$
that is $\mu(N)=0$, which contradicts the assumption that every open neighborhood of $x$ has positive measure. Therefore $x \in \operatorname{supp}(\mu)$. $\square$

Regularity Property

Theorem 1. Suppose that $\mu$ is a finite Borel measure on a metrizable space $(X,\tau)$. Then $\mu$ is closed regular, that is, if $A\in \mathcal{B}$ then

$$
\mu(A)=\sup\{\mu(B); B \text{ closed},\ B\subseteq A\}=\inf\{\mu(C); C \text{ open},\ C\supseteq A\}
$$
Proof : Let $d$ be a metric on $X$ which defines the topology, and let

$$
T=\{A\in \mathcal{B}\ ;\ \mu(A)=\sup\{\mu(B); B \text{ closed},\ B\subseteq A\}=\inf\{\mu(C); C \text{ open},\ C\supseteq A\} \}
$$

Suppose first that $A$ is closed, and that
$$
U_j=\{x\in X:d(x,A)<1/j.\}
$$
Then
$$
U_j\supseteq U_{j+1}
$$
and
$$
\bigcap_{j=1}^\infty U_j=A.
$$

Hence
$$
\mu(A)=\lim_{j\to\infty}\mu(U_j)=\inf_{j\in \mathbb{N}}\mu(U_j)\ge \inf\{\mu(C); C \text{ open},\ C\supseteq A\}
$$
Since for all $C$ open and $C\supseteq A$, we have $\mu(C)\ge \mu(A)$, then
$$
\inf\{\mu(C); C \text{ open},\ C\supseteq A\}\ge \mu(A).
$$
Therefore
$$
\mu(A)=\inf\{\mu(C); C \text{ open},\ C\supseteq A\}.
$$
Since $A$ is closed,
$$
\mu(A)=\sup\{\mu(B); B \text{ closed},\ B\subseteq A\}
$$
is trivial, we have $A\in T$.

It is therefore enough to show that $T$ is a $\sigma$-field. First, $\varnothing, X\in T$ is trivial. Next, if $A\in T$, that is

$$
\mu(A)=\sup\{\mu(B); B \text{ closed},\ B\subseteq A\}=\inf\{\mu(C); C \text{ open},\ C\supseteq A\}.
$$

Then
$$
\begin{aligned}
\mu(X\setminus A)&=\mu(X)-\mu(A)\\\
&=\mu(X)-\sup\{\mu(B); B \text{ closed},\ B\subseteq A\}\\\
&=\inf\{\mu(X\setminus B ); B \text{ closed},\ B\subseteq A\}\\\
&= \inf\{\mu(C); C \text{ open},\ C\supseteq A\},
\end{aligned}
$$
and similarly,
$$
\begin{aligned}
\mu(X\setminus A)&=\mu(X)-\mu(A)\\\
&=\mu(X)-\inf\{\mu(C); C \text{ open},\ C\supseteq A\}\\\
&=\sup\{\mu(X\setminus C ); C \text{ open},\ C\supseteq A\}\\\
&=\sup\{\mu(B); B \text{ closed},\ B\subseteq A\}.
\end{aligned}
$$

Therefore, $X\setminus A\in T$. Finally, it is enough to show that if $A_n,\ n=1,2,\cdots$ is a sequence in $T$, then
$$
A=\bigcup_{n=1}^{\infty} A_n\in T.
$$

Suppose that $\varepsilon>0$. Then for each $n$ there exist $F_n\subseteq A_n\subseteq U_n$ $(F_n$ closed, $U_n$ open$)$ with
$$
\mu(A_n\setminus F_n)<\frac{\varepsilon}{2^{n+1}}\quad \text{and}\quad \mu(U_n\setminus A_n)<\frac{\varepsilon}{2^{n}}.
$$

Then
$$
U=\bigcup_{n=1}^\infty U_n
$$
is open, and
$$
U\setminus A\subseteq \bigcup_{n=1}^{\infty} (U_n\setminus A_n),
$$

so that
$$
\mu(U\setminus A)\leq \sum_{n=1}^{\infty} \mu(U_n\setminus A_n)<\varepsilon.
$$

Let
$$
B_n=\bigcup_{i=1}^n A_i.
$$

Then $B_n\uparrow A$, and so there exists $N$ such that
$$
\mu(A\setminus B_N)<\frac{\varepsilon}{2}.
$$

Then
$$
G_N=\bigcup_{i=1}^N F_j
$$

is closed, and
$$
B_N\setminus G_N\subseteq \bigcup_{i=1}^N (A_i\setminus F_i),
$$

so that

$$
\mu(B_N\setminus G_N)\leq \sum_{i=1}^N \mu(A_i\setminus F_i)<\varepsilon/2.
$$

Thus
$$
\mu(A\setminus G_N)<\varepsilon.
$$
Therefore $A\in T$. $\square$

Reference

Chapter 16 in the following reference:

Garling, David JH. Analysis on Polish spaces and an introduction to optimal transportation. Vol. 89. Cambridge University Press, 2018.

The cover image of this article was taken at the Q1 Building, Gold Coast, Australia.