Weak Convergence Through Bounded Continuous Tests

Why Test Functions Describe Convergence of Laws

A probability measure records how probability is distributed across a space. To compare two such measures, one can ask whether they assign similar probabilities to particular events. But event-by-event comparisons can be delicate: small changes in a random vector may change the probability of an event whose boundary is heavily concentrated under the limiting law.

A different approach is to compare averages of functions. A bounded continuous function changes gradually as its input changes, and its integral summarizes how the measure is distributed across the space. Requiring these integrals to converge gives a central notion of convergence for probability measures: weak convergence.

Definition on Euclidean Space

Write $\mathcal{B}(\mathbb{R}^d)$ for the Borel sets of Euclidean space $\mathbb{R}^d$. A Borel probability measure $\mu$ assigns a probability to each set in $\mathcal{B}(\mathbb{R}^d)$. For a measurable function $f$ that is integrable with respect to $\mu$, its integral is denoted by $\int f\,d\mu$.

Definition: A sequence $(\mu_n)_{n\in\mathbb{N}}$ of Borel probability measures on $\mathbb{R}^d$ converges weakly to a Borel probability measure $\mu$ if, for every bounded continuous function $f:\mathbb{R}^d\to\mathbb{R}$, the integrals of $f$ converge: $$ \int_{\mathbb{R}^d} f(x)\,\mu_n(dx) \longrightarrow \int_{\mathbb{R}^d} f(x)\,\mu(dx). $$ We write $\mu_n\Rightarrow\mu$.

The test functions are required to be both bounded and continuous. Continuity makes their values responsive to small changes in the input; boundedness ensures that each integral is finite and prevents a test from giving arbitrarily large weight to rare, distant outcomes.

Laws Turn Random Vectors into Measures

If $X$ is a random vector taking values in $\mathbb{R}^d$, its law, denoted by $\mathcal{L}(X)$, is the probability measure defined by $\mathcal{L}(X)(B)=\mathbb{P}(X\in B)$ for every $B\in\mathcal{B}(\mathbb{R}^d)$. The random vector may be defined on any probability space; its law is a measure on $\mathbb{R}^d$.

Proposition: If $X$ is an $\mathbb{R}^d$-valued random vector and $f:\mathbb{R}^d\to\mathbb{R}$ is bounded and Borel measurable, then $$ \int_{\mathbb{R}^d} f(x)\,\mathcal{L}(X)(dx) = \mathbb{E}[f(X)]. $$

Proof. First take $f=\mathbf{1}_B$, the indicator of a Borel set $B$. The definition of the law gives $\int \mathbf{1}_B\,d\mathcal{L}(X)=\mathcal{L}(X)(B)=\mathbb{P}(X\in B)=\mathbb{E}[\mathbf{1}_B(X)]$. By linearity, the identity follows for every Borel simple function, which is a finite linear combination of indicators.

For a bounded nonnegative Borel function $f$, there is a sequence of nonnegative Borel simple functions increasing pointwise to $f$. The monotone convergence theorem applied to both sides gives the identity for $f$. A bounded real-valued Borel function is the difference of its bounded nonnegative parts $f^+=\max(f,0)$ and $f^-=\max(-f,0)$. Applying the nonnegative case to each part and subtracting proves the identity for $f$. $\square$

Consequently, if $X_n$ and $X$ are random vectors, possibly defined on different probability spaces, then their laws converge weakly exactly when $\mathbb{E}[f(X_n)]\to\mathbb{E}[f(X)]$ for every bounded continuous $f:\mathbb{R}^d\to\mathbb{R}$. This is the same definition expressed in the language of random vectors.

Worked Examples

Worked Example: Translated Gaussian Laws

Let $Z$ be a real standard normal random variable, and let $(a_n)$ be a real sequence with $a_n\to a$. Set $X_n=Z+a_n$ and $X=Z+a$. We show that $\mathcal{L}(X_n)\Rightarrow\mathcal{L}(X)$.

Take any bounded continuous $f:\mathbb{R}\to\mathbb{R}$. For every outcome at which $Z$ is finite, $Z+a_n\to Z+a$, so continuity gives $f(Z+a_n)\to f(Z+a)$. Also, $|f(Z+a_n)|\leq\|f\|_\infty$, where $\|f\|_\infty=\sup_{x\in\mathbb{R}}|f(x)|<\infty$. The dominated convergence theorem therefore gives

\int f\,d\mathcal{L}(X_n) =\mathbb{E}[f(Z+a_n)] \longrightarrow \mathbb{E}[f(Z+a)] =\int f\,d\mathcal{L}(X).

This holds for every bounded continuous $f$, so the claimed weak convergence follows. No density calculation is needed.

Worked Example: A Vector Converging to a Diagonal

Let $U$ be uniformly distributed on $[0,1]$. Define $X_n=(U,U+1/n)$ and $X=(U,U)$, which are random vectors in $\mathbb{R}^2$. For each outcome, the Euclidean distance between these vectors is

\|X_n-X\|_2 = \sqrt{(U-U)^2+(U+1/n-U)^2} = \frac{1}{n}.

Thus $X_n\to X$ for every outcome. If $f:\mathbb{R}^2\to\mathbb{R}$ is bounded and continuous, then $f(X_n)\to f(X)$ for every outcome and $|f(X_n)|\leq\|f\|_\infty$. Dominated convergence and the law-integral identity yield

\int f\,d\mathcal{L}(X_n) =\mathbb{E}[f(X_n)] \longrightarrow \mathbb{E}[f(X)] =\int f\,d\mathcal{L}(X).

Hence $\mathcal{L}(X_n)\Rightarrow\mathcal{L}(X)$. The limit law is supported on the diagonal $\{(u,u):0\leq u\leq1\}$, even though each $X_n$ takes values on a different line segment.

Worked Example: A Rare Tail Can Preserve a Different Mean

For each $n\in\mathbb{N}$ with $n\geq2$, let $\mu_n$ be the mixture that equals the standard normal law with probability $1-1/n$ and the normal law with mean $n$ and variance $1$ with probability $1/n$. Let $\mu$ be the standard normal law. For any bounded continuous $f:\mathbb{R}\to\mathbb{R}$, write $Z$ for a standard normal random variable. The mixture definition gives

\int f\,d\mu_n-\int f\,d\mu = \frac{1}{n}\left(\mathbb{E}[f(Z+n)]-\mathbb{E}[f(Z)]\right).

Each expectation on the right has absolute value at most $\|f\|_\infty$. Therefore

\left|\int f\,d\mu_n-\int f\,d\mu\right| \leq \frac{2\|f\|_\infty}{n} \longrightarrow 0.

Thus $\mu_n\Rightarrow\mu$. But the means do not converge to the mean of $\mu$: the standard normal component has mean $0$, while the other component has mean $n$, so

\int x\,\mu_n(dx) = \left(1-\frac{1}{n}\right)0+\frac{1}{n}n =1, \qquad \int x\,\mu(dx)=0.

The function $f(x)=x$ that detects this difference is continuous but unbounded, so it is not among the tests required by weak convergence.

A Useful Link to Almost-Sure Convergence

Theorem: Suppose $X_n$ and $X$ are random vectors in $\mathbb{R}^d$, defined on the same probability space, and $X_n\to X$ almost surely. Then $\mathcal{L}(X_n)\Rightarrow\mathcal{L}(X)$.

Proof. Let $f:\mathbb{R}^d\to\mathbb{R}$ be any bounded continuous function. On the event of probability one where $X_n\to X$, continuity implies $f(X_n)\to f(X)$. Boundedness gives $|f(X_n)|\leq\|f\|_\infty$, an integrable constant on a probability space. The dominated convergence theorem and the law-integral identity now show

\int f\,d\mathcal{L}(X_n) = \mathbb{E}[f(X_n)] \longrightarrow \mathbb{E}[f(X)] = \int f\,d\mathcal{L}(X).

Since this holds for every bounded continuous $f$, the definition gives $\mathcal{L}(X_n)\Rightarrow\mathcal{L}(X)$. $\square$

The theorem is useful when random vectors are naturally constructed on a common probability space: almost-sure convergence of the vectors immediately supplies convergence of their laws. The definition itself, however, concerns only the laws. It does not require the random vectors to share a probability space or to be related outcome by outcome.

What Weak Convergence Does—and Does Not—Control

Weak convergence compares averages against a specified class of tests. It does not say that every function has convergent integrals, nor does it directly assert convergence of probabilities for every Borel set. The rare-tail example shows why the boundedness condition matters: a small amount of probability can move far enough to change an unbounded average while having a vanishing effect on every fixed bounded test.

A common mistake is to infer convergence of expectations merely from convergence of laws. For a bounded continuous function, convergence of expectations is exactly what weak convergence guarantees. For an unbounded function, additional conditions are needed; in applications involving expectations, uniform integrability is one important tool for controlling contributions from large values.

The central practical procedure is simple: identify a bounded continuous test $f$, express its integral under each law as an expectation when random vectors are available, and prove that these expectations converge. To establish weak convergence, the argument must work for every such test, not just for a few convenient functions.

Check Your Understanding

Use the definition and examples above to check that you can distinguish convergence of laws from stronger forms of convergence.

State the definition of weak convergence for probability measures on $\mathbb{R}^d$. Which two properties must every test function have?
Why does $\int f\,d\mathcal{L}(X)=\mathbb{E}[f(X)]$ hold for bounded Borel $f$?
In the translated Gaussian example, where are continuity and boundedness used?
Why do the rare-tail measures converge weakly even though their means remain equal to $1$, rather than converging to $0$?
What does the almost-sure convergence theorem imply about the laws of random vectors on a shared probability space?