# The completeness assumption in Prokhorov's theorem

Originally, I encountered this question on Terence Tao's blog, where the following exercise is presented:

Exercise 23 (Implications and equivalences) Let $X_n, X$ be random variables taking values in a $\sigma$compact metric space $R$.

[...]

(ii) Show that if $X_n$ converges in distribution to $X$, then $X_n$ has a tight sequence of distributions.

(iii) Show that if $X_n$ converges in probability to $X$, then $X_n$ converges in distribution to $X$. (Hint: first show tightness, then use the fact that on compact sets, continuous functions are uniformly continuous.)

Whats struck me as odd was that $R$ is merely assumed to be $\sigma$-compact, i.e. no completeness-assumption on $R$ is made (as the example $\Bbb{Q}$ shows, there are $\sigma$-compact spaces that are neither locally compact, nor complete).

This makes it rather hard to construct new compact subsets (from old ones). Indeed, the proofs of the above statement (ii) that I found (see e.g. https://www.math.leidenuniv.nl/~vangaans/jancol1.pdf Theorem 5.2) use the fact that

$$K := \bigcap_j \bigcup_{i=1}^{k_j} \overline{B}(a_i, 1/j)$$ is a compact set if $R$ is complete, because it is closed and totally bounded.

Nevertheless, Wikipedia (http://en.wikipedia.org/wiki/Prokhorov%27s_theorem) does also not assume that the metric space in question is complete, there the only requirement is separability.

Note that we need to produce compact subsets of $R$, because for tightness (see below), we have to show that for $\varepsilon > 0$ there is $K \subset R$ compact such that $\Bbb{P}(X_n \in K) \geq 1-\varepsilon$ holds for all $n$ sufficiently large.

In summary, my question is if the two statements as in the exercise above are correct even without further completeness assumptions on $R$. Hints/proofs/counterexamples would be highly appreciated.

For convenience of the reader, I repeat the necessary definitions below:

Definition 10 (Modes of convergence) Let $R = (R,d)$ be a $\sigma$-compact metric space (with the Borel $\sigma$-algebra), and let $X_n$ be a sequence of random variables taking values in $R$. Let $X$ be another random variable taking values in $R$.

[...]

• $X_n$ converges in probability to $X$ if, for every $\epsilon > 0$, one has $$\liminf_{n \rightarrow \infty} {\bf P}( d(X_n,X) \leq \epsilon ) = 1$$ [...]

• $X_n$ converges in distribution to $X$ if, for every bounded continuous function $F: R \rightarrow {\bf R}$, one has $$\lim_{n \rightarrow\infty} \mathop{\bf E} F(X_n) = \mathop{\bf E} F(X)$$

• $X_n$ has a tight sequence of distributions if, for every $\epsilon > 0$, there exists a compact subset $K$ of $R$ such that $\mathop{\bf P}( X_n \in K ) \geq 1 - \epsilon$ for all sufficiently large $n$.

• Note that Exercise 23 makes no compactness claims, just tightness claims.
– Otis
Jun 21, 2014 at 21:36
• Yes, but tightness means that there is a compact set $K$ such that $\Bbb{P}(X_n \in K) \geq 1-\varepsilon$ holds for all $n$ large enough. I will edit the question to make it more clear. Jun 21, 2014 at 21:46

The assumption of $\sigma$-compactness is used to guarantee that each probability measure is tight. In this case, there is no need of completeness or to use the fact that a probability Borel measure on a Polish space is tight because if $X=\bigcup_n K_n$ where each $K_n$ is compact and $K_n\subset K_{n+1}$, then $\mu(K_n)\uparrow 1$.

In order to prove (ii), fix $\varepsilon$ and $n$ such that $\mathbb P(X\in K_n)\gt 1-\varepsilon$. Then consider a continuous function $F$ such that $F(x)=1$ if $x\in K_n$, $F(x)=0$ if $x\in K_{n+1}^c$ and $0\leqslant F\leqslant 1$. Using the definition of convergence in distribution, we obtain the wanted result.

Statement (iii) is also true and it can be shown using portmanteau theorem.

• Thank you very much for your thoughts. There is only one problem: In general there is no such function $F$ (it would in particular have compact support $\subset K_{n+1}$). Consider for example $R=\Bbb{Q}$. Then every continuous function $F$ with compact support vanishes identically (otherwise, $B_\varepsilon (x) \cap \Bbb{Q} \subset \rm{supp}(F) \subset \Bbb{Q}$, which entails that $\rm{supp}(F)$ can not be compact). EDIT: As I just noted, this implies in particular, that part (iv) of the exercise fails in general). Jun 22, 2014 at 19:08
• You are right (maybe it will work if we only assume $F$ non-negative, bounded and $f(x)=1$ if $x\in K$). Jun 22, 2014 at 19:19

As it turns out, the claim is indeed true.

The $\sigma$-compactness of the space $R$ ensures that $R$ is still $\sigma$-compact (and thus a Borel-set) in the completion of $R$. Also (which turns out to be equivalent (for separable metric spaces)) it ensures that every finite measure on $R$ is tight.

These two (for separable metric spaces equivalent) conditions are also called universal measurability of $R$, cf. Dudley, Real Analysis and Probability, Theorem 11.5.1 and the definition before that theorem.

Then a result of Le Cam (cf. Dudley, Theorem 11.5.3) implies that if $X_n \rightarrow X$ in distribution, then the sequence of laws $\mathcal{L}(X_n)$ is uniformly tight, which is exactly what I wanted to know.

• Dudley's books deals with the case when $R$ is $\sigma$-compact. Do you know whether separability is enough (as in the statement on Wikipedia?) Aug 13, 2022 at 23:53