Radon Measure

Definition 1. Suppose that $f$ is a function on the Borel subsets of a metric space $X$ taking values in $[0,\infty]$. We say $f$ is locally finite if for each $x \in X$ there exists a neighborhood $N$ of $x$ with $f(N) < \infty$.

Proposition 1. If $f:\mathcal{B}(X)\to [0,\infty]$ is a locally finite additive function on a metrizable space $(X,\tau)$ and $K$ is a compact subset of $X$ then $f(K) < \infty$.

Proof: For each $x \in K$ there exists a neighborhood $N_x$ of $x$ with $f(N_x) < \infty$. The sets $\{N_x:x\in K\}$ cover $K$, and so there is a finite subcover. Additivity then ensures that $f(K) < \infty$. $\square$

Definition 2. A Radon measure $\mu$ on a metrizable space $(X,\tau)$ is a tight additive function from $\mathcal{B}(X)$ to $[0,\infty]$ which is locally finite.

Remark: Since non-negative additive tight function on the Borel sets of a metrizable space $(X,\tau)$ is also $\sigma$-additive, a Radon measure $\mu$ is indeed a measure, and, by the preceding property, $\mu(K) < \infty$ if $K$ is compact.

Proposition 2. If $\mu$ is a Radon measure on a separable metrizable space, then $\mu$ is $\sigma$-finite; there exists a countable set $\mathcal{W}$ of open sets for which
$$
X=\bigcup_{W\in\mathcal{W}}W
$$
and $f(W)<\infty$, for all $W\in\mathcal{W}$.

Proof: Let $d$ be a metric on $X$ which defines the topology $\tau$. For each $n \in \mathbb{N}$, let

$$
U_n=\{x\in X:\text{there exists }r_x>1/n\text{ such that }f(N_{r_x}(x))<\infty\}.
$$

First, we show that $U_n$ is open for each $n\in \mathbb{N}$. For $x\in U_n$ and
$$
d(x,y)< r_x-\frac{1}{n}.
$$
Then let
$$
s_y=r_x-d(x,y)>r_x-(r_x-\frac{1}{n})=\frac{1}{n}.
$$
Now we claim that

$$
N_{s_y}(y)\subseteq N_{r_x}(x).
$$

Actually, for all $z\in N_{s_y}(y)$, we have
$$
d(z,x)\le d(z,y)+d(y,x)< s_y+d(y,x)=r_x,
$$
which means $z\in N_{r_x}(x)$. Therefore, $N_{s_y}(y)\subseteq N_{r_x}(x)$. Hence $U_n$ is open.

Let $C_n$ be a countable dense subset of $U_n$, and let

$$
W_n(c)=N_{1/n}(c)\quad \text{ for } c\in C_n.
$$

Then $f(W_n(c))<\infty$ for each $c\in C_n$ due to Radon measure $\mu$ is locally finite.

If $x\in U_n$ then there exists $c\in C_n$ with $d(x,c)<1/n$, so that

$$
U_n\subseteq \bigcup_{c\in C_n}W_n(c).
$$

Let $\mathcal{W}=\{W_n(c):n\in \mathbb{N},,c\in C_n\}$, then

$$
X=\bigcup_{W\in\mathcal{W}}W.\quad \square
$$

Remark: Suppose that $X$ and $Y$ are metric spaces, that $f:X\to Y$ is continuous and that $\mu$ is a Radon measure on $X$. Then the push-forward measure $f_\star(\mu)$ need not be a Radon measure on $Y$.

Counterexample: Let $\mu$ be counting measure on $\mathbb{N}$, and let $f:\mathbb{N}\to [0,\infty]$ be the inclusion mapping. Then $f_*(\mu)$ is not locally finite at $\infty$.

Theorem 1. Let $(X,\tau)$ be a Polish space, and let $(U_i)_{i=1}^\infty$ be a sequence of open subsets of $X$ such that

$$
X=\bigcup_{i=1}^\infty U_i.
$$

Suppose that for each $i$, $\mu_i$ is a finite measure on the Borel sets of $U_i$ and that these measures are compatible: if $A$ is a Borel set of $U_i\cap U_j$, then

$$
\mu_i(A)=\mu_j(A).
$$

Then there exists a unique Radon measure $\pi$ on $X$ for which

$$
\pi(A)=\mu_i(A)
$$

for each Borel set $A$ in $U_i$, for each $i\in \mathbb{N}$.

Proof: Let

$$
V_j=\bigcup_{i=1}^j U_i.
$$

The compatibility condition ensures that we can define a finite positive Borel measure $\nu_j$ on $V_j$ such that

$$
\nu_j(A)=\mu_i(A)
$$

if $1\le i\le j$ and $A\subseteq U_i$ is Borel.

Further, if $A$ is a Borel subset of $V_j$ and $j\le k$, then

$$
\nu_j(A)=\nu_k(A).
$$

If $A$ is a Borel subset of $X$, then $(\nu_j(A\cap V_j))_{j=1}^\infty$ is an increasing sequence. Indeed, if $j\le k$, then

$$
A\cap V_j\subseteq A\cap V_k,
$$

and since $\nu_k$ extends $\nu_j$ on $V_j$,

$$
\nu_j(A\cap V_j)=\nu_k(A\cap V_j)\le \nu_k(A\cap V_k).
$$

Let

$$
\pi(A)=\lim_{j\to\infty}\nu_j(A\cap V_j).
$$

We now verify that $\pi$ is tight, locally finite and additive.

First, $\pi$ is locally finite. Let $x\in X$. Since $\bigcup_{i=1}^\infty U_i=X$, there exists $i$ such that $x\in U_i$. Because $U_i\subseteq V_j$ for every $j\ge i$, if $A\subseteq U_i$ is Borel then

$$
\nu_j(A)=\mu_i(A), \qquad j\ge i.
$$

Hence

$$
\pi(A)=\lim_{j\to\infty}\nu_j(A\cap V_j)=\mu_i(A).
$$

Taking $A=U_i$, we obtain

$$
\pi(U_i)=\mu_i(U_i)<\infty.
$$

Thus $x$ has an open neighbourhood of finite $\pi$-measure, and $\pi$ is locally finite.

Next, $\pi$ is additive. Let $A,B\subseteq X$ be disjoint Borel sets. Then for each $j$,

$$
(A\cup B)\cap V_j=(A\cap V_j)\cup (B\cap V_j),
$$

and the two sets on the right-hand side are disjoint. Since $\nu_j$ is a measure,

$$
\nu_j((A\cup B)\cap V_j)=\nu_j(A\cap V_j)+\nu_j(B\cap V_j).
$$

Passing to the limit gives

$$
\pi(A\cup B)=\pi(A)+\pi(B).
$$

So $\pi$ is additive.

Finally, $\pi$ is tight. Let $A$ be a Borel subset of $X$ with $\pi(A)<\infty$, and let $\varepsilon>0$. By definition of $\pi(A)$, there exists $j$ such that

$$
\pi(A)-\nu_j(A\cap V_j)<\varepsilon/2.
$$

Since $V_j$ is an open subspace of the Polish space $X$, it is itself Polish. The measure $\nu_j$ is a finite Borel measure on the Polish space $V_j$, hence it is a Radon measure on $V_j$. Therefore there exists a compact set $K\subseteq A\cap V_j$ such that

$$
\nu_j((A\cap V_j)\setminus K)<\varepsilon/2.
$$

Because $K\subseteq V_j$, we have

$$
\pi(K)=\nu_j(K).
$$

Also,

$$
A\setminus K=((A\cap V_j)\setminus K)\cup (A\setminus V_j),
$$

and these two sets are disjoint. Hence, by additivity,

$$
\pi(A\setminus K)=\pi((A\cap V_j)\setminus K)+\pi(A\setminus V_j).
$$

Now

$$
\pi((A\cap V_j)\setminus K)=\nu_j((A\cap V_j)\setminus K)<\varepsilon/2,
$$

and

$$
\pi(A\setminus V_j)=\pi(A)-\pi(A\cap V_j)=\pi(A)-\nu_j(A\cap V_j)<\varepsilon/2.
$$

Therefore

$$
\pi(A\setminus K)<\varepsilon.
$$

So $\pi$ is tight.

Thus $\pi$ is a tight, locally finite additive function on the Borel subsets of the metrizable space $X$. Hence $\pi$ is a Radon measure.

Moreover, as already shown above, if $A$ is a Borel subset of $U_i$, then for every $j\ge i$,

$$
\nu_j(A)=\mu_i(A),
$$

so

$$
\pi(A)=\mu_i(A).
$$

Thus $\pi$ has the required restriction property.

It remains to prove uniqueness. Let $\widetilde{\pi}$ be another Radon measure on $X$ such that

$$
\widetilde{\pi}(A)=\mu_i(A)
$$

for every Borel set $A\subseteq U_i$ and every $i$. Then for each $j$, the restrictions of $\widetilde{\pi}$ and $\nu_j$ to every $U_i$, $1\le i\le j$, coincide. Since $V_j=\bigcup_{i=1}^j U_i$, it follows that

$$
\widetilde{\pi}(B)=\nu_j(B)
$$

for every Borel set $B\subseteq V_j$.

Now let $A\subseteq X$ be Borel. Since

$$
A\cap V_1\subseteq A\cap V_2\subseteq \cdots
\quad\text{and}\quad
\bigcup_{j=1}^\infty (A\cap V_j)=A,
$$

and since $\widetilde{\pi}$ is a measure, continuity from below gives

$$
\widetilde{\pi}(A)=\lim_{j\to\infty}\widetilde{\pi}(A\cap V_j).
$$

But for each $j$,

$$
\widetilde{\pi}(A\cap V_j)=\nu_j(A\cap V_j).
$$

Therefore

$$
\widetilde{\pi}(A)=\lim_{j\to\infty}\nu_j(A\cap V_j)=\pi(A).
$$

So $\widetilde{\pi}=\pi$, and the proof is complete. $\square$

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 in Lucerne, Switzerland.