Proving the Continuous Mapping Theorem

Why the Proof Is a Test-Function Argument

Weak convergence is defined by what happens to expectations of bounded continuous functions. A continuous mapping theorem therefore has a natural proof strategy: take an arbitrary bounded continuous test function on the transformed space, compose it with the map, and use the definition of weak convergence on the original space.

The important point is that the composition remains a bounded continuous test function. Continuity follows from continuity of both functions, and boundedness comes from the test function—not from any boundedness assumption on the map. This makes the argument work even when the transformed random vectors can take arbitrarily large values.

Weak Convergence and Laws of Transformed Vectors

Let \(X_n\) and \(X\) be random vectors taking values in \(\mathbb{R}^d\); they need not be defined on the same probability space. Write \(\mathcal{L}(X_n)\) and \(\mathcal{L}(X)\) for their probability laws. Weak convergence of \(X_n\) to \(X\) means weak convergence of these laws.

Definition: The sequence \(X_n\) converges weakly to \(X\), written \(X_n \Rightarrow X\), if for every bounded continuous function \(h:\mathbb{R}^d\to\mathbb{R}\),

\mathbb{E}[h(X_n)]\longrightarrow \mathbb{E}[h(X)].

If \(f:\mathbb{R}^d\to\mathbb{R}^k\) is Borel measurable, the law of \(f(X)\) is the image, or pushforward, of the law of \(X\) under \(f\). In particular, for a Borel set \(A\subseteq\mathbb{R}^k\),

\mathcal{L}(f(X))(A) = \mathbb{P}(f(X)\in A) = \mathcal{L}(X)(f^{-1}(A)).

A continuous map between Euclidean spaces is Borel measurable, so this law is well defined whenever \(f\) is continuous.

The Test-Function Identity for a Pushforward Law

The proof will use an integral identity that connects a transformed random vector with its original one. If \(P\) is a probability measure on \(\mathbb{R}^d\), and \(f:\mathbb{R}^d\to\mathbb{R}^k\) is Borel measurable, write \(f_{\#}P\) for the pushforward measure, defined by \(f_{\#}P(A)=P(f^{-1}(A))\) for Borel \(A\).

Proposition: For every bounded Borel function \(g:\mathbb{R}^k\to\mathbb{R}\),

\int_{\mathbb{R}^k} g(y)\,(f_{\#}P)(dy) = \int_{\mathbb{R}^d} g(f(x))\,P(dx).

Proof. First take \(g\) to be the indicator of a Borel set \(A\). By the definition of the pushforward measure,

\(\int \mathbf{1}_A(y)\,(f_{\#}P)(dy)=(f_{\#}P)(A)=P(f^{-1}(A))=\int \mathbf{1}_A(f(x))\,P(dx)\).

Linearity gives the identity for nonnegative simple functions. Every nonnegative Borel function is the increasing pointwise limit of nonnegative simple functions, so the monotone convergence theorem gives the identity for nonnegative Borel functions. For bounded real-valued \(g\), apply that result to its positive and negative parts and subtract the resulting finite integrals. This proves the stated identity.

\(\square\)

The Continuous Mapping Theorem

Theorem: Suppose \(X_n\Rightarrow X\) in \(\mathbb{R}^d\), and \(f:\mathbb{R}^d\to\mathbb{R}^k\) is continuous everywhere. Then \(f(X_n)\Rightarrow f(X)\) in \(\mathbb{R}^k\).

Proof. Let \(g:\mathbb{R}^k\to\mathbb{R}\) be any bounded continuous function. Since \(f\) is continuous, the composition \(g\circ f:\mathbb{R}^d\to\mathbb{R}\) is continuous. Since \(g\) is bounded, there is a finite constant \(M\) such that \(|g(y)|\leq M\) for every \(y\in\mathbb{R}^k\). Consequently, \(|g(f(x))|\leq M\) for every \(x\in\mathbb{R}^d\), whether or not \(f\) is bounded. Thus \(g\circ f\) is a bounded continuous test function on \(\mathbb{R}^d\).

By \(X_n\Rightarrow X\), the defining test-function condition applies to \(g\circ f\):

\mathbb{E}[g(f(X_n))] = \mathbb{E}[(g\circ f)(X_n)] \longrightarrow \mathbb{E}[(g\circ f)(X)] = \mathbb{E}[g(f(X))].

This holds for every bounded continuous \(g:\mathbb{R}^k\to\mathbb{R}\). By the definition of weak convergence on \(\mathbb{R}^k\), it follows that \(f(X_n)\Rightarrow f(X)\).

\(\square\)

Equivalently, if \(P_n=\mathcal{L}(X_n)\) and \(P=\mathcal{L}(X)\), the conclusion says that \(f_{\#}P_n\) converges weakly to \(f_{\#}P\). The proposition explains the same proof at the level of measures: each test integral against a pushforward law is an integral of the composed test against the original law.

Worked Examples

Worked Example: Squaring a Weakly Convergent Sequence

Suppose \(X_n\Rightarrow X\) in \(\mathbb{R}\). Take \(f(x)=x^2\). This map is continuous on all of \(\mathbb{R}\), since it is a polynomial. To verify the weak-convergence conclusion directly, let \(g:\mathbb{R}\to\mathbb{R}\) be bounded and continuous. Then \(x\mapsto g(x^2)\) is continuous as a composition, and it is bounded by the same bound as \(g\). Therefore,

\mathbb{E}[g(X_n^2)] = \mathbb{E}[(g\circ f)(X_n)] \longrightarrow \mathbb{E}[(g\circ f)(X)] = \mathbb{E}[g(X^2)].

As this holds for every such \(g\), \(X_n^2\Rightarrow X^2\). No assumption that the \(X_n\) or \(X\) are bounded is needed.

Worked Example: Transforming a Joint Vector by Addition

Suppose \((U_n,V_n)\Rightarrow(U,V)\) in \(\mathbb{R}^2\). Define \(f:\mathbb{R}^2\to\mathbb{R}\) by \(f(x,y)=x+y\). This map is continuous: if \((x_m,y_m)\to(x,y)\), then \(x_m+y_m\to x+y\). For an arbitrary bounded continuous \(g:\mathbb{R}\to\mathbb{R}\), the composition is

(g\circ f)(x,y)=g(x+y).

It is continuous and bounded on \(\mathbb{R}^2\). Applying the definition of joint weak convergence gives

\mathbb{E}[g(U_n+V_n)] \longrightarrow \mathbb{E}[g(U+V)].

Hence \(U_n+V_n\Rightarrow U+V\). The assumption here is convergence of the joint vectors, not merely separate weak convergence of \(U_n\) and \(V_n\).

Worked Example: A Vector-Valued Polynomial Transformation

Suppose \((X_n,Y_n)\Rightarrow(X,Y)\) in \(\mathbb{R}^2\), and define \(f:\mathbb{R}^2\to\mathbb{R}^2\) by \(f(x,y)=(x+y,xy)\). Both coordinates are polynomials, so \(f\) is continuous. Let \(g:\mathbb{R}^2\to\mathbb{R}\) be any bounded continuous test function. Its composition with \(f\) is

(g\circ f)(x,y)=g(x+y,xy).

This is continuous and bounded, because it is the composition of continuous maps and its values are values of the bounded function \(g\). Thus

\mathbb{E}[g(X_n+Y_n,X_nY_n)] \longrightarrow \mathbb{E}[g(X+Y,XY)].

Since the convergence holds for every bounded continuous \(g\) on \(\mathbb{R}^2\), the theorem yields \((X_n+Y_n,X_nY_n)\Rightarrow(X+Y,XY)\). The output is a vector, so the test functions used to verify its weak convergence must be defined on \(\mathbb{R}^2\).

Why Boundedness of the Map Is Not Required

A common concern is that a continuous map such as \(f(x)=x^2\) or \(f(x)=e^x\) can take very large values, while weak convergence is tested using bounded functions. There is no conflict. The test is \(g\), defined on the output space; it is bounded by assumption. Composing it with \(f\) cannot increase the range beyond the range of \(g\). In particular, the proof does not require \(\mathbb{E}|f(X_n)|\) to be finite.

This also shows why the theorem is specifically suited to weak convergence. Its definition already asks for convergence of expectations of every bounded continuous test, and the composition supplies exactly another test of that kind. No separate estimate of moments or tails is involved.

Scope of the Result

The hypothesis in this version is continuity of \(f\) at every point of the domain. It guarantees that \(g\circ f\) is continuous everywhere, which is precisely what the test-function definition requires. One should not replace this hypothesis casually with measurability: measurability makes \(f(X_n)\) a valid random vector, but it does not ensure that composing a continuous test with \(f\) produces a continuous test.

The result transfers an already established weak limit through a fixed continuous transformation. Its proof does not assert that separate coordinate limits automatically give joint convergence; the addition example required joint convergence as its starting assumption. Nor does the proof require all random vectors to share a probability space: it compares their laws through expectations, so each vector may be defined on its own probability space.

Key takeaway: To prove weak convergence after a continuous transformation, test the transformed vectors with an arbitrary bounded continuous function \(g\). The composed function \(g\circ f\) is bounded and continuous, so the original weak convergence applies directly.

Check Your Understanding

Use the test-function definition in your answers, and identify the domain and range of each map.

In the proof of the theorem, which assumption ensures that \(g\circ f\) is continuous?
Why is \(g\circ f\) bounded even if \(f\) is unbounded?
If \(X_n\Rightarrow X\) in \(\mathbb{R}\), what does the theorem imply for the map \(f(x)=x^2\)?
For the addition example, why is joint weak convergence of \((U_n,V_n)\) the stated starting assumption?
State the pushforward integral identity for a bounded Borel function \(g\), and explain how it relates to the theorem’s proof.