Bounded-Lipschitz Tests Determine Weak Convergence

Why a Smaller Class of Tests Can Suffice

Weak convergence is defined by testing integrals against every bounded continuous function. That class is broad: a continuous function need not have a finite Lipschitz constant, even on all of Euclidean space. It is therefore useful to know whether a more manageable class of tests can still detect weak convergence.

For probability measures on \(\mathbb{R}^d\), bounded Lipschitz functions do suffice. The key is not that every bounded continuous function is globally well approximated by a Lipschitz function. Instead, convergence of the tests first gives control of the measures outside a large ball. On the ball, continuity provides the approximation that is needed.

Definitions and the Criterion

Definition: A function \(f:\mathbb{R}^d\to\mathbb{R}\) is Lipschitz if there is a finite constant \(L\geq 0\) such that \(|f(x)-f(y)|\leq L\|x-y\|\) for all \(x,y\in\mathbb{R}^d\). It is bounded Lipschitz if it is both bounded and Lipschitz. Here \(\|\cdot\|\) is the Euclidean norm.

Let \(\mu_n\) and \(\mu\) be Borel probability measures on \(\mathbb{R}^d\). Weak convergence, written \(\mu_n\Rightarrow\mu\), means that the integrals converge for every bounded continuous real-valued function \(g\):

\int g\,d\mu_n\longrightarrow\int g\,d\mu.

In the theorem below, convergence “for every bounded Lipschitz function” is pointwise in the test function: each such function is fixed while \(n\) tends to infinity. No single bound on the Lipschitz constants of all test functions is required.

Theorem (bounded-Lipschitz criterion): For Borel probability measures \(\mu_n,\mu\) on \(\mathbb{R}^d\), \(\mu_n\Rightarrow\mu\) if and only if

\int f\,d\mu_n\longrightarrow\int f\,d\mu \quad\text{for every bounded Lipschitz } f:\mathbb{R}^d\to\mathbb{R}.

The forward implication is immediate: a bounded Lipschitz function is bounded and continuous, so it is among the tests in the definition of weak convergence. The reverse implication is the substantive part. We establish it by controlling tails and then approximating on a compact set.

Two Ingredients for the Converse

First, the limiting probability measure puts arbitrarily little mass outside sufficiently large balls. Write \(B_R=\{x\in\mathbb{R}^d:\|x\|\leq R\}\). Since these balls increase to \(\mathbb{R}^d\), countable additivity gives \(\mu(B_R)\to 1\) as \(R\to\infty\). A Lipschitz cutoff turns this fact into a tail bound for \(\mu_n\) as well.

Tail-control argument: Fix \(\eta>0\) and choose \(R\) such that \(\mu(B_R)>1-\eta\). Define

\psi_R(x)=\min\{1,\max\{0,\|x\|-R\}\}.

The norm is 1-Lipschitz, and clipping a real-valued function to the interval \([0,1]\) does not increase its Lipschitz constant. Thus \(\psi_R\) is bounded Lipschitz. It vanishes on \(B_R\), equals 1 when \(\|x\|\geq R+1\), and satisfies

\int\psi_R\,d\mu\leq \mu(B_R^c)<\eta.

If the integrals of all bounded Lipschitz functions converge, then eventually \(\int\psi_R\,d\mu_n<2\eta\). Because \(\psi_R=1\) outside \(B_{R+1}\), this yields

\mu_n(B_{R+1}^c)\leq\int\psi_R\,d\mu_n<2\eta

for all sufficiently large \(n\). This is the tail control needed below; no bound on moments is being assumed.

The second ingredient is uniform approximation on a compact set. We include the construction because it explains exactly where compactness enters.

Lemma: Let \(K\subseteq\mathbb{R}^d\) be compact, and let \(g:\mathbb{R}^d\to\mathbb{R}\) be bounded and continuous. For every \(\varepsilon>0\), there is a bounded Lipschitz function \(h:\mathbb{R}^d\to\mathbb{R}\) such that \(|h(x)-g(x)|<\varepsilon\) for every \(x\in K\).

Proof. Set \(C=\sup_{x\in\mathbb{R}^d}|g(x)|\). If \(C=0\), take \(h=0\). Otherwise, continuity of \(g\) on the compact set \(K\) implies uniform continuity there. Choose \(\delta>0\) such that \(x,y\in K\) and \(\|x-y\|<\delta\) imply \(|g(x)-g(y)|<\varepsilon\). Choose \(L>0\) with \(L\delta\geq 2C\), and set

q(x)=\inf_{y\in K}\bigl(g(y)+L\|x-y\|\bigr).

For any \(x,z\in\mathbb{R}^d\), the triangle inequality gives \(\|x-y\|\leq\|z-y\|+\|x-z\|\). Taking infima over \(y\in K\) shows \(q(x)\leq q(z)+L\|x-z\|\); interchanging \(x,z\) proves that \(q\) is \(L\)-Lipschitz. The infimum is finite because \(K\) is nonempty and compact and \(g\) is bounded there. If \(K\) is empty, the conclusion is vacuous and any bounded Lipschitz \(h\) works.

For \(x\in K\), choosing \(y=x\) gives \(q(x)\leq g(x)\). For any \(y\in K\) with \(\|x-y\|<\delta\), uniform continuity gives \(g(y)+L\|x-y\|\geq g(x)-\varepsilon\). If instead \(\|x-y\|\geq\delta\), then

g(y)+L\|x-y\|\geq -C+L\delta\geq C\geq g(x).

Consequently \(g(x)-\varepsilon\leq q(x)\leq g(x)\) on \(K\). Finally, clip \(q\) to the interval \([-C,C]\), defining \(h(x)=\max\{-C,\min\{C,q(x)\}\}\). Clipping is 1-Lipschitz as a function of a real variable, so \(h\) is bounded Lipschitz. Since \(g(x)\in[-C,C]\) on \(K\), clipping preserves the stated approximation there. This proves the lemma.

\square

Proof of the Bounded-Lipschitz Criterion

Proof. Assume that the integrals converge for every bounded Lipschitz function, and fix an arbitrary bounded continuous \(g:\mathbb{R}^d\to\mathbb{R}\). We prove convergence of its integrals.

If \(g\) is identically zero, the conclusion is immediate. Otherwise, let \(C=\sup_x|g(x)|>0\). Fix \(\eta>0\). By the tail-control argument, choose \(R\) so that \(\mu(B_R)>1-\eta\), and set \(K=B_{R+1}\). For all sufficiently large \(n\), \(\mu_n(K^c)<2\eta\), while \(\mu(K^c)<\eta\). By the lemma, for any \(\varepsilon>0\) there is a bounded Lipschitz function \(h\) with \(|h-g|<\varepsilon\) on \(K\); the construction also ensures \(|h|\leq C\) everywhere.

Split the integral difference into its contributions from \(K\) and \(K^c\). On \(K\), \(|h-g|<\varepsilon\); on \(K^c\), both functions have absolute value at most \(C\). Therefore, for all sufficiently large \(n\),

\begin{aligned} \left|\int g\,d\mu_n-\int g\,d\mu\right| &\leq \left|\int h\,d\mu_n-\int h\,d\mu\right|\\ &\quad+\left|\int(g-h)\,d\mu_n\right| +\left|\int(h-g)\,d\mu\right|\\ &\leq \left|\int h\,d\mu_n-\int h\,d\mu\right| +2\varepsilon+4C\eta+2C\eta. \end{aligned}

For the fixed \(h\), the first term tends to zero by the assumed convergence of bounded Lipschitz tests. Hence the limit superior of the left-hand side is at most \(2\varepsilon+6C\eta\). Since \(\varepsilon\) and \(\eta\) can be chosen arbitrarily small, this limit superior is zero. Thus the integrals converge for the arbitrary bounded continuous \(g\), proving \(\mu_n\Rightarrow\mu\).

Conversely, if \(\mu_n\Rightarrow\mu\), every bounded Lipschitz function is bounded and continuous, so its integrals converge by the definition of weak convergence. Both implications hold.

\square

Worked Examples

Worked Example: Convergence of Point Masses

Let \(x_n,x\in\mathbb{R}^d\). Then \(\delta_{x_n}\Rightarrow\delta_x\) if and only if \(x_n\to x\). If \(x_n\to x\), every bounded Lipschitz \(f\) is continuous, so \(f(x_n)\to f(x)\), which is exactly convergence of its integrals against these point masses. Conversely, if the point masses converge weakly, apply the bounded-Lipschitz criterion to \(f(z)=\min\{1,\|z-x\|\}\). This function is bounded and 1-Lipschitz, and

\int f\,d\delta_{x_n}=f(x_n)=\min\{1,\|x_n-x\|\} \longrightarrow f(x)=0.

For any \(0<\rho<1\), eventually \(\min\{1,\|x_n-x\|\}<\rho\), which implies \(\|x_n-x\|<\rho\). Taking \(\rho\) arbitrarily small gives \(x_n\to x\).

Worked Example: Rare Mass Escaping to Infinity

Let \(e_1=(1,0,\ldots,0)\), and define \(\mu_n=(1-1/n)\delta_0+(1/n)\delta_{ne_1}\) for \(n\in\mathbb{N}\). For any bounded continuous \(f\),

\left|\int f\,d\mu_n-f(0)\right| =\frac{1}{n}|f(ne_1)-f(0)| \leq\frac{2\sup_x|f(x)|}{n}\longrightarrow 0.

Thus \(\mu_n\Rightarrow\delta_0\), and in particular every bounded Lipschitz test has convergent integrals. Yet the first moment does not converge to the first moment of \(\delta_0\):

\int\|x\|\,d\mu_n(x) =\left(1-\frac1n\right)0+\frac1n\|ne_1\|=1, \qquad \int\|x\|\,d\delta_0(x)=0.

There is no contradiction: the function \(x\mapsto\|x\|\) is unbounded and is not one of the tests in the definition of weak convergence.

Worked Example: A Test Detects a Persistent Discrepancy

Consider \(\mu_n=\delta_0\) for even \(n\) and \(\mu_n=\delta_{e_1}\) for odd \(n\). The bounded Lipschitz function \(f(x)=\min\{1,\|x\|\}\) satisfies

\int f\,d\mu_n= \begin{cases} 0,&n\text{ even},\\ 1,&n\text{ odd}. \end{cases}

This sequence of integrals does not converge, so the measures cannot converge weakly. The example illustrates why a proposed weak limit must pass every bounded Lipschitz test, not merely a few tests that happen to give stable values.

What the Criterion Does—and Does Not—Say

The proof has two distinct parts. The cutoff functions prevent mass from escaping unnoticed when passing from a compact region to all of \(\mathbb{R}^d\). The compact-set approximation then replaces an arbitrary bounded continuous test by a bounded Lipschitz one where the measures have nearly all their mass. Omitting tail control would leave the approximation argument incomplete: approximation on a fixed ball says nothing about what happens outside that ball.

The criterion concerns convergence of probability measures, not convergence of integrals for unbounded functions. The rare-mass example shows why this distinction matters. Nor does the theorem say that one particular Lipschitz test is enough. Its conclusion relies on convergence for every bounded Lipschitz function.

Check Your Understanding

Use the definitions and proof to check your understanding of how the test class characterizes weak convergence.

Why does weak convergence immediately imply convergence of integrals for bounded Lipschitz functions?
What values does the cutoff \(\psi_R\) take on \(B_R\) and outside \(B_{R+1}\), and how does this give a tail bound?
In the approximation lemma, why is continuity on a compact set enough, even though \(g\) need not be uniformly continuous on all of \(\mathbb{R}^d\)?
Why must the proof control the measures outside the compact set before using the approximation lemma?
In the rare-mass example, why does weak convergence not force the first moments to converge?