The Continuous Mapping Theorem for Weak Limits

Why Transforming a Weak Limit Matters

Weak convergence describes the limiting behavior of probability measures through bounded continuous test functions. In applications, however, the quantities of interest are often transformed: a measurement may be squared, a vector summarized by its norm, or a scalar encoded together with a nonlinear function of itself. The continuous mapping theorem gives a direct way to transfer a weak limit through such transformations.

The theorem is especially useful because it operates at the level of probability laws. One need not compute the transformed distributions explicitly. It is enough to know that the original laws converge weakly and that the map being applied is continuous everywhere.

Pushforward Measures

Let \(\mu\) be a probability measure on \(\mathbb{R}^d\), and let \(f:\mathbb{R}^d\to\mathbb{R}^k\) be a Borel measurable map. The map transports probability from the original space to the target space: a target set receives the probability of its preimage.

Definition: The pushforward measure of \(\mu\) under \(f\), denoted \(f_{\#}\mu\), is the probability measure on \(\mathbb{R}^k\) defined, for each Borel set \(B\subseteq\mathbb{R}^k\), by \[ (f_{\#}\mu)(B)=\mu(f^{-1}(B)). \]

The inverse image \(f^{-1}(B)\) is Borel because \(f\) is measurable. In particular, \(f^{-1}(\mathbb{R}^k)=\mathbb{R}^d\), so the pushforward assigns total mass one. If \(X\) is a random vector with law \(\mu\), then \(f_{\#}\mu\) is the law of \(f(X)\).

Proposition: If \(f:\mathbb{R}^d\to\mathbb{R}^k\) is Borel measurable, then for every nonnegative Borel function \(h:\mathbb{R}^k\to[0,\infty]\), \[ \int_{\mathbb{R}^k}h(y)\,(f_{\#}\mu)(dy) = \int_{\mathbb{R}^d}h(f(x))\,\mu(dx). \] The identity also holds for every bounded Borel function \(h\).

Proof. First let \(h=\mathbf{1}_B\) for a Borel set \(B\subseteq\mathbb{R}^k\). The left-hand side is \((f_{\#}\mu)(B)\), which equals \(\mu(f^{-1}(B))\) by definition. The right-hand side is \(\int \mathbf{1}_{f^{-1}(B)}\,d\mu\), the same quantity.

By linearity, the identity therefore holds for every nonnegative Borel simple function. For a general nonnegative Borel \(h\), choose a sequence of nonnegative Borel simple functions \(h_m\) increasing pointwise to \(h\). The identity holds for each \(h_m\). The monotone convergence theorem on both sides gives the stated identity for \(h\). A bounded Borel \(h\) is the difference of its nonnegative positive and negative parts, both of which are bounded; applying the nonnegative case to each part proves the bounded case.

\(\square\)

This identity explains why pushforwards are a natural language for transformed laws: integrating a function of the transformed value is equivalent to integrating its composition with the map in the original space.

The Continuous Mapping Theorem

For probability measures on Euclidean spaces, weak convergence can be expressed by convergence of integrals against every bounded continuous real-valued function. The continuous mapping theorem says that a continuous transformation preserves this convergence when applied to the measures.

Theorem (Continuous Mapping Theorem): Let \((\mu_n)_{n\in\mathbb{N}}\) and \(\mu\) be probability measures on \(\mathbb{R}^d\), and suppose \(\mu_n\) converges weakly to \(\mu\). Let \(f:\mathbb{R}^d\to\mathbb{R}^k\) be continuous at every point. Then \[ f_{\#}\mu_n \Rightarrow f_{\#}\mu. \]

Here \(\Rightarrow\) denotes weak convergence. In random-vector language, if \(X_n\) has law \(\mu_n\) and \(X\) has law \(\mu\), the conclusion is that \(f(X_n)\) converges weakly to \(f(X)\). This concerns their probability laws; it does not require the random vectors to be defined on the same probability space.

The theorem applies to a continuous map on the entire source space. The map need not be bounded. For a vector-valued transformation, the domain and target dimensions can differ; only continuity and the stated weak convergence of the original laws are required.

Worked Examples

Worked Example: Squaring a Rarely Large Observation

For \(n\in\mathbb{N}\), define the probability measure on \(\mathbb{R}\) \[ \mu_n=\left(1-\frac{1}{n}\right)\delta_0+\frac{1}{n}\delta_n, \] where \(\delta_a\) is the point mass at \(a\). For any bounded continuous \(\varphi:\mathbb{R}\to\mathbb{R}\), \[ \int \varphi\,d\mu_n-\varphi(0) =\frac{\varphi(n)-\varphi(0)}{n}. \] Thus the absolute difference is at most \(2\|\varphi\|_{\infty}/n\), which tends to zero. Hence \(\mu_n\Rightarrow\delta_0\).

Take \(f(x)=x^2\), which is continuous on \(\mathbb{R}\). Its pushforward is \[ f_{\#}\mu_n=\left(1-\frac{1}{n}\right)\delta_0+\frac{1}{n}\delta_{n^2}, \] so the continuous mapping theorem gives \(f_{\#}\mu_n\Rightarrow\delta_0=f_{\#}\delta_0\). Directly, for bounded continuous \(\psi\), the difference from \(\psi(0)\) is \((\psi(n^2)-\psi(0))/n\), again bounded in absolute value by \(2\|\psi\|_{\infty}/n\).

Although the transformed laws converge weakly, their means do not converge to the mean of the limiting law: \(\int x\,d(f_{\#}\mu_n)=n\cdot(1/n)=1\), while \(\int x\,d\delta_0=0\). The function \(x\mapsto x\) is unbounded, so this expectation comparison is not a consequence of weak convergence.

Worked Example: A Scalar and Its Exponential Jointly

Suppose \(\mu_n\Rightarrow\mu\) on \(\mathbb{R}\), and let \(f(x)=(x,e^x)\), a continuous map from \(\mathbb{R}\) to \(\mathbb{R}^2\). The theorem gives \[ f_{\#}\mu_n\Rightarrow f_{\#}\mu. \] Equivalently, if \(X_n\) has law \(\mu_n\) and \(X\) has law \(\mu\), then \((X_n,e^{X_n})\Rightarrow(X,e^X)\).

The limit law is supported on the curve \(\{(x,e^x):x\in\mathbb{R}\}\). To see the test-function content of the conclusion, take any bounded continuous \(\Phi:\mathbb{R}^2\to\mathbb{R}\). The function \(x\mapsto\Phi(x,e^x)\) is bounded and continuous. The pushforward integral identity gives \[ \int_{\mathbb{R}^2}\Phi\,d(f_{\#}\mu_n) =\int_{\mathbb{R}}\Phi(x,e^x)\,\mu_n(dx), \] and the theorem asserts convergence of these transformed laws to \(f_{\#}\mu\). The two coordinates in the limit are linked by the same input \(X\); the statement does not treat them as independent.

Worked Example: Applying a Norm to Random Vectors

Let \(\mu_n\Rightarrow\mu\) on \(\mathbb{R}^2\), and choose the Euclidean norm map \(f(x,y)=\sqrt{x^2+y^2}\). This map is continuous from \(\mathbb{R}^2\) to \(\mathbb{R}\), so \[ f_{\#}\mu_n\Rightarrow f_{\#}\mu. \] Thus if \(Z_n\) has law \(\mu_n\) and \(Z\) has law \(\mu\), then \(\|Z_n\|\Rightarrow\|Z\|\).

For a concrete check, take \(\mu_n=\delta_{(3,\,4+1/n)}\) and \(\mu=\delta_{(3,4)}\). For every bounded continuous \(\Phi:\mathbb{R}^2\to\mathbb{R}\), \(\int\Phi\,d\mu_n=\Phi(3,4+1/n)\to\Phi(3,4)=\int\Phi\,d\mu\), so \(\mu_n\Rightarrow\mu\). The image measures are \[ f_{\#}\mu_n=\delta_{\sqrt{9+(4+1/n)^2}}, \qquad f_{\#}\mu=\delta_5. \] Since \(9+(4+1/n)^2=25+8/n+1/n^2\), the displayed point locations tend to \(5\), in agreement with weak convergence of the image measures.

Worked Example: An Affine Transformation of a Vector

Let \(A\) be a fixed \(k\times d\) real matrix and \(b\in\mathbb{R}^k\), and suppose \(\mu_n\Rightarrow\mu\) on \(\mathbb{R}^d\). The map \(f(x)=Ax+b\) is continuous, so \(f_{\#}\mu_n\Rightarrow f_{\#}\mu\). In particular, if \(X_n\) and \(X\) have the corresponding laws, then \(AX_n+b\Rightarrow AX+b\).

This covers coordinate rescaling, translation, and linear combinations as instances of one theorem. For example, with \(d=2\), \(k=1\), \(A=(2\;\;-3)\), and \(b=1\), the map is \(f(x_1,x_2)=2x_1-3x_2+1\). Its continuity follows because it is a finite linear combination of coordinate functions and a constant. No calculation of the transformed density or mass function is needed.

Composing Transformations

Pushforwards also behave consistently when transformations are applied in sequence. This fact lets one express a chain of continuous operations either as successive pushforwards or as a single pushforward under the composite map.

Proposition: Let \(f:\mathbb{R}^d\to\mathbb{R}^k\) and \(g:\mathbb{R}^k\to\mathbb{R}^{\ell}\) be Borel measurable. For every probability measure \(\mu\) on \(\mathbb{R}^d\), \[ (g\circ f)_{\#}\mu=g_{\#}(f_{\#}\mu). \]

Proof. For any Borel set \(C\subseteq\mathbb{R}^{\ell}\), the definitions give \[ \bigl(g_{\#}(f_{\#}\mu)\bigr)(C) =(f_{\#}\mu)(g^{-1}(C)) =\mu\bigl(f^{-1}(g^{-1}(C))\bigr). \] The inverse-image identity \(f^{-1}(g^{-1}(C))=(g\circ f)^{-1}(C)\) then yields \[ \bigl(g_{\#}(f_{\#}\mu)\bigr)(C) =\mu\bigl((g\circ f)^{-1}(C)\bigr) =\bigl((g\circ f)_{\#}\mu\bigr)(C). \] The two measures agree on every Borel set \(C\), so they are equal.

\(\square\)

For example, one may first map a vector to its norm and then apply a continuous scalar function to that norm. The composition is continuous, and the theorem can be applied directly to the composite map; the proposition confirms that the resulting law is the same as the law obtained by carrying out the transformations one at a time.

What the Theorem Does—and Does Not—Give

The useful feature of the theorem is that it transfers a limit of laws without requiring explicit formulas for those laws. Once the original weak limit is known, a continuous transformation produces the corresponding transformed limit. This makes the theorem a standard tool for identifying limiting distributions of summaries and reparameterizations.

A common mistake is to infer convergence of expectations for every function from weak convergence. Weak convergence controls bounded continuous test functions. A transformation \(f\) itself may be unbounded, as the squaring example illustrates, and expectations of unbounded quantities can behave differently. Additional hypotheses are needed for conclusions about such expectations.

Another point to keep precise is the role of continuity. The stated theorem assumes continuity at every point of the source space. Measurability alone ensures that a pushforward can be defined, but it is not the continuity hypothesis in this version of the theorem. When applying it, specify the domain and target of the map and verify that the map is continuous on the whole domain.

Check Your Understanding

Use the definitions and theorem above to answer the following questions.

For a Borel measurable \(f\), how is the probability of a target set under \(f_{\#}\mu\) defined?
If \(\mu_n\Rightarrow\mu\) on \(\mathbb{R}^d\) and \(f:\mathbb{R}^d\to\mathbb{R}^k\) is continuous everywhere, what is the weak limit of \(f_{\#}\mu_n\)?
Why does the continuous mapping theorem not, by itself, imply convergence of the means in the rare-event squaring example?
Suppose \(f\) and \(g\) are measurable maps with compatible domains. What identity relates \((g\circ f)_{\#}\mu\) to successive pushforwards?
If \(X_n\Rightarrow X\) and \(f\) is continuous everywhere, what does the theorem say about the laws of \(f(X_n)\) and \(f(X)\)?