When Is the Space of Probability Measures Compact?

The answer depends on the topology placed on the set of measures. The standard interpretation is the weak topology, in which probability measures are close when they give nearly the same integrals to bounded continuous functions. Under this topology, compactness is governed by tightness: probability mass must not be able to escape to infinity.

The compact-space theorem

Let $X$ be a compact metric space, and let $\mathcal{P}(X)$ denote its Borel probability measures. Then $\mathcal{P}(X)$ is compact in the weak topology. In particular, every sequence of probability measures on $X$ has a weakly convergent subsequence, and its limit is again a probability measure on $X$.

Weak convergence means that $\mu_n$ converges to $\mu$ if

$$\int_X f\,d\mu_n \longrightarrow \int_X f\,d\mu$$

for every bounded continuous function $f:X\to\mathbb{R}$. On a compact space, every continuous real-valued function is bounded, so one can equivalently test against all continuous functions.

A compactness proof can be understood through Prokhorov’s theorem: any collection of probability measures on compact $X$ is tight, because the single compact set $X$ itself contains all the mass. Tightness then gives relative compactness. The space $\mathcal{P}(X)$ is also closed among probability measures, so the full space is compact.

The general criterion: Prokhorov’s theorem

A metric space is Polish if its metric can be chosen complete and it has a countable dense subset. On a Polish space $X$, Prokhorov’s theorem says that a family $\mathcal{F}\subseteq\mathcal{P}(X)$ is relatively compact in the weak topology if and only if it is uniformly tight. Uniform tightness means that for every $\varepsilon>0$, there is a compact set $K\subseteq X$ such that

$$\mu(K)\geq 1-\varepsilon \qquad\text{for every }\mu\in\mathcal{F}.$$

Thus, to determine whether a particular family is relatively compact, check whether one compact set can capture almost all the mass for every measure in the family. “Relatively compact” means its closure is compact; it does not necessarily mean the family itself is closed or compact.

For the entire space $\mathcal{P}(X)$, this criterion gives a useful exact result: if $X$ is Polish, then $\mathcal{P}(X)$ is weakly compact if and only if $X$ is compact. If $X$ is compact, tightness is immediate. If $X$ is not compact, the point-mass measures $\delta_x$ cannot be uniformly tight: for any compact $K\subset X$, choose $x\notin K$, and then $\delta_x(K)=0$.

Worked examples

On $X=[0,1]$, consider $\mu_n=\delta_{1/n}$, the probability measure concentrated at $1/n$. For any continuous $f:[0,1]\to\mathbb{R}$,

$$\int f\,d\mu_n=f(1/n)\longrightarrow f(0)=\int f\,d\delta_0.$$

Therefore $\delta_{1/n}$ converges weakly to $\delta_0$. This illustrates how a sequence of measures can converge by testing their integrals, and the compact-space theorem guarantees that every sequence in $\mathcal{P}([0,1])$ has some weakly convergent subsequence.

By contrast, on $\mathbb{R}$ consider $\nu_n=\delta_n$. No compact subset of $\mathbb{R}$ contains all sufficiently large integers, so for any compact $K$, $\nu_n(K)=0$ for all sufficiently large $n$. The family is not tight, and it has no weakly convergent subsequence in $\mathcal{P}(\mathbb{R})$. The mass is escaping to infinity.

Common mistakes

• Leaving out the topology. Compactness is not a property of the set of measures alone. The result above concerns the weak topology.
• Confusing weak convergence with total-variation convergence. Even when $X$ is compact, $\mathcal{P}(X)$ is generally not compact in total variation. For distinct points $x_n$, the measures $\delta_{x_n}$ remain separated in total variation, so they cannot have a convergent subsequence there.
• Treating tightness as compactness by itself. Prokhorov’s theorem applies to families of measures on Polish spaces and yields relative compactness. To conclude compactness of a family, also check that it is closed.
• Assuming boundedness prevents escape. On a noncompact space, measures can move their mass into regions no fixed compact set captures, as $\delta_n$ does on $\mathbb{R}$.

So, when someone asks whether “the space of probability measures” is compact, first identify the underlying space and the topology. With weak convergence, compact metric underlying spaces give compact spaces of probability measures; on Polish spaces, tightness is the test for relative compactness.