Weak convergence of a sequence probability measures, denoted $P_n \Rightarrow P$, is typically defined as,

$$ \int\limits_{\mathbb{R}}fdP_n \xrightarrow{n\rightarrow\infty} \int\limits_{\mathbb{R}}fdP \;\;\;\; (*)$$

for all $f \in C_b(\mathbb{R})$, the space of continuous and bounded functions. However I read here that if $(*)$ holds for all $f \in C_c^\infty(\mathbb{R})$, the space of smooth compactly supported functions, then weak convergence also holds. The proof in the link is as follows:

Fix $\epsilon > 0$ and choose a smooth compactly supported function $g$ with $0 \leq g \leq 1$ and $\int gdP \geq 1−ϵ$. Let $K$ be the support of $g$. Then by assumption, $$\int gdP_n\rightarrow\int gdP \geq 1 - \epsilon$$ so we see that, $P_n(K)\geq 1−\epsilon$. In other words, $\{P_n\}$ is tight. Hence after passing to a subsequence, $\{P_{n_k}\}$ converges weakly to some measure $P^*$. Now we notice that $\int fdP =\int fdP^*$ for all $f\in C_c^\infty(\mathbb{R})$, and it follows from a monotone class type argument that $P = P^*$. Finally use the "double subsequence" trick to conclude that the original sequence $\{P_n\}$ also converges weakly to $P$.

I think I am able to follow the above until the last sentence. How we can conclude weak convergence of the original sequence? All we've shown is that for all $f \in C_b(\mathbb{R})$,

$$ \int f dP_{n_k} \xrightarrow{k\rightarrow\infty} \int f dP $$

for this particular subsequence $\{P_{n_k}\}$. I'm not sure what the "double subsequence" trick is. Is it some sort of diagonal argument? How can we use it to conclude that,

$$ \int\limits_{\mathbb{R}}fdP_n \xrightarrow{n\rightarrow\infty} \int\limits_{\mathbb{R}}fdP $$

for all $f \in C_b(\mathbb{R})$?


1 Answer 1


The trick is also known as subsequence principle. For a sequence of real numbers it reads as follows:

Let $(a_n)_{n \in \mathbb{N}} \subseteq \mathbb{R}$ be a sequence. Then $a_n$ converges to some $a \in \mathbb{R}$ if, and only if the following statement holds: For any subsequence of $(a_n)_{n \in \mathbb{N}}$ there exists a further subsequence which converges, and the limit does not depend on the chosen subsequence.

In your setting this principle is used for weak convergence.

The proof (the one which you stated in your question) shows that any subsequence of $(P_n)_n$ has a subsequence which converges weakly to $P$. Now suppose that $P_n$ does not converge weakly to $P$. Then we can find a subsequence $(P_{n_k})_k$ such that

$$\left| \int g \, dP_{n_k} - \int g \, dP \right| > \epsilon \quad \text{for all $k \geq 1$}$$

for some $\epsilon>0$ and a suitable function $g$. In particular, any subsequence of $(P_{n_k})_k$ does not converge weakly to $P$ which is a contradiction.

  • 1
    $\begingroup$ Ah, so my understanding of Prokhorov's theorem is not correct. If the sequence of $\{P_n\}$ is tight, then any subsequence of $\{P_n\}$ has a further subsequence that converges weakly to some $P$? I thought the theorem stated that there exists a subsequence of the original $\{P_n\}$ that converges weakly to $P$. $\endgroup$
    – user460426
    Feb 4, 2018 at 21:33
  • 1
    $\begingroup$ @Flowsnake It is obvious that any subsequence $(P_{n_k})_k$ is tight, and therefore you can apply Prokhorov's theorem to this sequence to find a subsequence of $(P_{n_k})_k$ which converges weakly to $P$. $\endgroup$
    – saz
    Feb 5, 2018 at 8:22

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