Continuous mapping theorem

Proof

This proof has been adopted from (van der Vaart 1998, Theorem 2.3)

Spaces S and S′ are equipped with certain metrics. For simplicity we will denote both of these metrics using the |x − y| notation, even though the metrics may be arbitrary and not necessarily Euclidean.

Convergence in distribution

We will need a particular statement from the portmanteau theorem: that convergence in distribution ${\displaystyle X_n\stackrel{d}{\to }X}$ is equivalent to

${\displaystyle \mathbb{E}f(X_n)\to \mathbb{E}f(X)}$ for every bounded continuous functional f.

So it suffices to prove that ${\displaystyle \mathbb{E}f(g(X_n))\to \mathbb{E}f(g(X))}$ for every bounded continuous functional f. For simplicity we assume g continuous. Note that ${\displaystyle F=f\circ g}$ is itself a bounded continuous functional. And so the claim follows from the statement above. The general case is slightly more technical.

Convergence in probability

Fix an arbitrary ε > 0. Then for any δ > 0 consider the set B_δ defined as

${\displaystyle B_{\delta }=\{x\in S\mid x\notin D_g:\mathrm{\exists }y\in S:|x-y|<\delta ,|g(x)-g(y)|>\epsilon \}.}$

This is the set of continuity points x of the function g(·) for which it is possible to find, within the δ-neighborhood of x, a point which maps outside the ε-neighborhood of g(x). By definition of continuity, this set shrinks as δ goes to zero, so that lim_{δ → 0}B_δ = ∅.

Now suppose that |g(X) − g(X_n)| > ε. This implies that at least one of the following is true: either |X−X_n| ≥ δ, or X ∈ D_g, or X∈B_δ. In terms of probabilities this can be written as

${\displaystyle Pr(|g(X_n)-g(X)|>\epsilon )\le Pr(|X_n-X|\ge \delta )+Pr(X\in B_{\delta })+Pr(X\in D_g).}$

On the right-hand side, the first term converges to zero as n → ∞ for any fixed δ, by the definition of convergence in probability of the sequence {X_n}. The second term converges to zero as δ → 0, since the set B_δ shrinks to an empty set. And the last term is identically equal to zero by assumption of the theorem. Therefore, the conclusion is that

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}Pr(|g(X_n)-g(X)|>\epsilon )=0,}$

which means that g(X_n) converges to g(X) in probability.

Note: The above proof treats ${\displaystyle B_{\delta }}$ as a Borel subset of ${\displaystyle S}$, which is not clear in general. This can be circumvented easily as follows. By assumption, we have ${\displaystyle \mathbb{P}\left(\bigcup _{k\ge 1}{A}_k\right)=1}$, where ${\displaystyle A_k}$ is the event "for all ${\displaystyle n\ge 1}$ such that ${\displaystyle |X_n-X|\le 1/k}$, we have ${\displaystyle |g(X_n)-g(X)|<\epsilon }$":$${\displaystyle A_k=\bigcap _{n\ge 1}\left(|X_n-X|\le \frac{1}{k}\Rightarrow |g(X_n)-g(X)|<\epsilon \right).}$$Now, fix an arbitrary ${\displaystyle \eta >0}$. Since the sequence ${\displaystyle (A_k{)}_{k\ge 1}}$ is non-decreasing, we have ${\displaystyle \mathbb{P}\left(\bigcup _{k\ge 1}{A}_k\right)=\underset{k\to \mathrm{\infty }}{lim}\mathbb{P}(A_k)}$, hence there exists ${\displaystyle k\ge 1}$ such that ${\displaystyle \mathbb{P}(A_k)\ge 1-\eta }$. Finally, we can write

$${\displaystyle \mathbb{P}(|g(X_n)-g(X)|\ge \epsilon )\le \eta +\mathbb{P}(|g(X_n)-g(X)|\ge \epsilon ;A_k)\le \eta +\mathbb{P}\left(d(X_n,X)>\frac{1}{k}\right).}$$

It follows that ${\displaystyle \underset{n\to \mathrm{\infty }}{\overline{\lim }}\mathbb{P}(|g(X_n)-g(X)|\ge \epsilon )\le \eta +0=\eta }$, which shows that ${\displaystyle \mathbb{P}(|g(X_n)-g(X)|\ge \epsilon )\to 0}$ as ${\displaystyle n\to \mathrm{\infty }}$, since ${\displaystyle \eta }$ was arbitrary.

Almost sure convergence

By definition of the continuity of the function g(·),

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{X}_n(\omega )=X(\omega )\Rightarrow \underset{n\to \mathrm{\infty }}{lim}g(X_n(\omega ))=g(X(\omega ))}$

at each point X(ω) where g(·) is continuous. Therefore,

${\displaystyle \begin{array}{cc}Pr\left(\underset{n\to \mathrm{\infty }}{lim}g(X_n)=g(X)\right)& \ge Pr\left(\underset{n\to \mathrm{\infty }}{lim}g(X_n)=g(X),X\notin D_g\right)\\ & \ge Pr\left(\underset{n\to \mathrm{\infty }}{lim}{X}_n=X,X\notin D_g\right)=1,\end{array}}$

because the intersection of two almost sure events is almost sure.

By definition, we conclude that g(X_n) converges to g(X) almost surely.