Tightness of Borel Measure

Definition 1. A mapping $f$ from the Borel sets of a metrizable space $(X,\tau)$ to $[0,\infty]$ is tight if $f(K)<\infty$ for each compact $K$ in $X$ and
$$
f(A)=\sup\{f(K):K \text{ compact},\ K\subseteq A\},
\quad \text{for each } A\in \mathcal{B}(X).
$$

Tightness is very powerful, as the next result shows. We consider non-negative functions on the Borel subsets of a metric space $(X,\tau)$ which can take infinite values.

As before, a mapping $f:\mathcal{B}(X)\to [0,\infty]$ is additive if
$$
f(A\cup B)=f(A)+f(B),
$$
whenever $A$ and $B$ are disjoint, and is $\sigma$-additive if

$$
f!\left(\bigcup_{n=1}^{\infty} A_n\right)=\sum_{n=1}^{\infty} f(A_n)
$$

for each sequence $(A_n)_{n=1}^{\infty}$ of disjoint Borel sets.

Proposition 1. Suppose that $f$ is a non-negative additive tight function on the Borel sets of a metrizable space $(X,\tau)$. Then $f$ is $\sigma$-additive, and so it is a tight Borel measure on $X$.

Proof : Suppose that $(A_n)_{n=1}^{\infty}$ is a sequence of disjoint Borel sets whose union is $A$. First we consider the case where $f(A)<\infty$. Let

$$
B_n=\bigcup_{j=1}^n A_j.
$$

Since
$$
B_n\subseteq A,
$$

we have

$$
\sum_{j=1}^n f(A_j)\le f(A),
$$

and so

$$
\sum_{j=1}^{\infty} f(A_j)\le f(A).
$$

Suppose, if possible, that

$$
\sum_{j=1}^{\infty} f(A_j)=s<f(A).
$$

Let $\varepsilon=\frac{f(A)-s}{2}>0$ and $C_n=A\setminus B_n$. We have

$$
f(C_n)\ge f(A)-f(B_n)\ge f(A)-s= 2\varepsilon,
\quad \text{for all } n\in \mathbb{N}.
$$

By combining blocks of terms, we can suppose that

$$
f(B_n)>s-\frac{\varepsilon}{2^{n+2}},
\quad \text{for all } n\in \mathbb{N}.
$$

For each $n\in \mathbb{N}$ there exists a compact subset $K_n$ of $C_n$ with

$$
f(K_n)>f(C_n)-\frac{\varepsilon}{2^{n+1}}.
$$

Let

$$
L_n=\bigcap_{j=1}^n K_j.
$$

We now show by induction that

$$
f(L_n)\ge \left(1+\frac{1}{2^n}\right)\varepsilon
$$

so that

$$
f(C_n\setminus L_n)\le \left(1-\frac{1}{2^n}\right)\varepsilon.
$$

The result is true when $n=1$; suppose that it is true for $n$. Now

$$
\begin{aligned}
f(C_n\setminus C_{n+1})
&=f(B_{n+1}\setminus B_n) \\\
&=f(B_{n+1})-f(B_n) \\\
&\le s-\left(s-\frac{\varepsilon}{2^{n+2}}\right) \\\
&=\frac{\varepsilon}{2^{n+2}}.
\end{aligned}
$$

Since

$$
C_n\setminus K_{n+1}\subseteq (C_n\setminus C_{n+1})\cup (C_{n+1}\setminus K_{n+1}),
$$

it follows that

$$
f(C_n\setminus K_{n+1})\le f(C_n\setminus C_{n+1})+f(C_{n+1}\setminus K_{n+1})< \frac{\varepsilon}{2^{n+1}}.
$$

Since

$$
C_n=(C_n\setminus K_{n+1})\cup (C_n\setminus L_n)\cup (L_n\cap K_{n+1}),
$$

we have

$$
\begin{aligned}
2\varepsilon
&\le f(C_n) \\\
&\le f(C_n\setminus K_{n+1})+f(C_n\setminus L_n)+f(L_n\cap K_{n+1}) \\\
&\le \frac{\varepsilon}{2^{n+1}}+\left(1-\frac{1}{2^n}\right)\varepsilon+f(L_{n+1}),
\end{aligned}
$$

so that

$$
f(L_{n+1})\ge \left(1+\frac{1}{2^{n+1}}\right)\varepsilon.
$$

This establishes the induction.

But

$$
\bigcap_{n=1}^{\infty} L_n\subseteq \bigcap_{n=1}^{\infty} C_n=\varnothing.
$$

Since the sets $L_n$ are compact, it follows that there exists $N\in \mathbb{N}$ for which

$$
L_N=\varnothing,
$$

so that

$$
f(L_N)=0,
$$

giving a contradiction.

Finally, suppose that $f(A)=\infty$. If $M<\infty$, there exists a compact subset $K$ of $A$ with $f(K)>M$. Since $f$ is tight and by the definition of tightness, we have $f(K)<\infty$. Thus,

$$
\sum_{n=1}^{\infty} f(A_n)\ge \sum_{n=1}^{\infty} f(A_n\cap K)=f(K)>M,
$$

so that

$$
\sum_{n=1}^{\infty} f(A_n)=\infty.\quad \square
$$

Proposition 2. A finite Borel measure $\mu$ on a metric space $(X,d)$ is tight if and only if
$$
\sup\{\mu(K):K \text{ compact},\ K\subseteq X\}=\mu(X).
$$

Proof : The condition is certainly necessary. Suppose that it is satisfied. Suppose that $A$ is a Borel set, and that $\varepsilon>0$. There exists a closed set $B\subseteq A$ such that
$$
\mu(B)\ge \mu(A)-\varepsilon/2
$$

and there exists a compact $K$ such that

$$
\mu(K)>\mu(X)-\varepsilon/2.
$$

Then $B\cap K$ is a compact subset of $A$, and

$$
\begin{aligned}
\mu(B\cap K)&=\mu(B)-\mu(B\cap (X\setminus K))\\\
&\ge \mu(B)-\mu(X\setminus K)\\\
&\ge \mu(A)-\varepsilon/2-\varepsilon/2\\\
&= \mu(A)-\varepsilon.
\end{aligned}
$$

Therefore $\mu$ is tight. $\square$

Theorem 1(Ulam’s Theorem). A finite Borel measure on a Polish space $(X, \tau)$ is tight.

Proof : Let $d$ be a complete metric on $X$ which defines the topology $\tau$. Let $(c_j)_{j=1}^{\infty}$ be a dense sequence in $X$, and let

$$
M_{j,n} = \{ x \in X : d(x, c_j) \leq \frac{1}{n} \}
$$

and

$$
A_{j,n} = \bigcup_{i=1}^{j} M_{i,n}.
$$

Suppose that $\varepsilon > 0$. For $n \in \mathbb{N}$, each $A_{j,n}$ is closed, and $A_{j,n} \uparrow X$ as $j \to \infty$. Thus there exists $J_n$ such that if $E_n = A_{J_n,n}$, then
$$
\mu(E_n) > (1 - \frac{\varepsilon}{2^n}) \mu(X).
$$
It is easy to see that $E_n$ is totally bounded for each $n \in \mathbb{N}$. Let
$$
D_n = \bigcap_{j=1}^{n} E_j.
$$

Then $(D_n)_{n=1}^{\infty}$ is a decreasing sequence of closed sets, and for each $n \in \mathbb{N}$,

$$
\mu(D_n)=\mu(X)-\mu(\bigcup_{j=1}^{n} (X\setminus E_j))> \mu(X)-\sum_{j=1}^{n} \frac{\varepsilon}{2^j} \mu(X) = (1 - (1 - \frac{1}{2^n}) \varepsilon) \mu(X).
$$

Let
$$
D = \bigcap_{n=1}^{\infty} D_n,
$$

then
$$
\mu(D)=\lim_{n\to\infty}\mu(D_n) \geq (1 - \varepsilon) \mu(X).
$$

Moreover, $D$ is closed and totally bounded since $D\subseteq E_n$, and is therefore compact due to the space 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.