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Weak Convergence on Euclidean Spaces · Tutorial 162 of 946

Proving Portmanteau Bounds with Distance Functions

Use distance-to-set functions to prove the closed-set upper bound and open-set lower bound for probabilities under weak convergence.

Intermediate 10 min read

What You'll Learn

  • Define distance to a set and verify its continuity.
  • Construct continuous approximations to indicators of closed and open sets.
  • Prove both Portmanteau probability bounds from bounded-continuous-test convergence.
  • Track the order of limits in the approximation argument.
  • Use examples to see why weak convergence need not imply convergence of every event probability.

Why Approximate Set Indicators?

Weak convergence is defined through integrals of bounded continuous functions. But an event such as \(X_n\in F\) is represented by the indicator \(1_F\), which is generally not continuous. To control event probabilities using weak convergence, we need to connect these two kinds of functions.

In Euclidean space, distance provides a direct way to do this. A closed set can be approached from above by continuous functions that are at least its indicator. An open set can be approached from below by continuous functions bounded by its indicator. Those directions of approximation are what produce the upper and lower bounds.

Distance to a Set

For a nonempty set \(A\subseteq\mathbb{R}^d\), define the distance from \(x\in\mathbb{R}^d\) to \(A\) by

$$ d(x,A)=\inf_{y\in A}\|x-y\|. $$

The distance is always nonnegative. It is zero for \(x\in A\), and it can also be zero for points in the closure of \(A\). The function is continuous: in fact, it is Lipschitz with constant one.

Distance inequality: For every nonempty \(A\subseteq\mathbb{R}^d\) and every \(x,z\in\mathbb{R}^d\), \( |d(x,A)-d(z,A)|\leq \|x-z\| \).

To verify the inequality, fix \(y\in A\). The triangle inequality gives \(\|x-y\|\leq\|x-z\|+\|z-y\|\). Taking the infimum over \(y\in A\) yields \(d(x,A)\leq\|x-z\|+d(z,A)\). Interchanging \(x\) and \(z\) gives the claimed absolute-value bound. Thus distance functions are suitable building blocks for bounded continuous tests.

Approximating a Closed-Set Indicator from Above

Let \(F\subseteq\mathbb{R}^d\) be closed and nonempty. For each \(m\in\mathbb{N}\), define

$$ f_m(x)=\max\{1-m\,d(x,F),\,0\}. $$

The distance inequality shows that \(f_m\) is continuous. It takes values in \([0,1]\), and \(f_m(x)=1\) whenever \(x\in F\). Hence \(1_F(x)\leq f_m(x)\) for every \(x\). As \(m\) increases, \(f_m(x)\) decreases. If \(x\notin F\), closedness means that some open ball around \(x\) is disjoint from \(F\), so \(d(x,F)>0\); consequently \(f_m(x)=0\) for all sufficiently large \(m\). Therefore \(f_m\) decreases pointwise to \(1_F\).

Approximation from above: For closed \(F\), the bounded continuous functions \(f_m\) satisfy \(1_F\leq f_m\) and \(f_m\downarrow 1_F\) pointwise.

The Closed-Set Upper Bound

Suppose \((\mu_n)_{n\in\mathbb{N}}\) and \(\mu\) are probability measures on \(\mathbb{R}^d\), and \(\mu_n\Rightarrow\mu\). By definition, this weak convergence means that \(\int h\,d\mu_n\to\int h\,d\mu\) for every bounded continuous \(h:\mathbb{R}^d\to\mathbb{R}\).

Theorem: If \(\mu_n\Rightarrow\mu\) and \(F\subseteq\mathbb{R}^d\) is closed, then
$$ \limsup_{n\to\infty}\mu_n(F)\leq \mu(F). $$
Proof. If \(F=\varnothing\), both sides are zero. Otherwise, use the functions \(f_m\) defined above. Since \(1_F\leq f_m\), integration against \(\mu_n\) gives
$$ \mu_n(F)\leq \int f_m\,d\mu_n. $$

For each fixed \(m\), the function \(f_m\) is bounded and continuous, so weak convergence implies

$$ \limsup_{n\to\infty}\mu_n(F) \leq \int f_m\,d\mu. $$

This holds for every \(m\). Since \(f_m\downarrow 1_F\) and \(0\leq f_m\leq 1\), the dominated convergence theorem gives \(\int f_m\,d\mu\to\int 1_F\,d\mu=\mu(F)\) as \(m\to\infty\). Taking that limit in the bound proves the result.

\(\square\)

Approximating an Open-Set Indicator from Below

Let \(G\subseteq\mathbb{R}^d\) be open with nonempty complement. Define

$$ g_m(x)=\min\{1,m\,d(x,G^c)\}. $$

This is again bounded and continuous. It is zero outside \(G\), so \(g_m\leq 1_G\). If \(x\in G\), openness gives a ball around \(x\) contained in \(G\); therefore \(d(x,G^c)>0\), and \(g_m(x)\) eventually equals one. Thus \(g_m\) increases pointwise to \(1_G\). If \(G=\mathbb{R}^d\), its probability is always one, so the desired lower bound is immediate; if \(G=\varnothing\), it is immediate as well.

Approximation from below: For open \(G\) with nonempty complement, the functions \(g_m\) satisfy \(0\leq g_m\leq 1_G\) and \(g_m\uparrow 1_G\) pointwise.

The Open-Set Lower Bound

Theorem: If \(\mu_n\Rightarrow\mu\) and \(G\subseteq\mathbb{R}^d\) is open, then
$$ \liminf_{n\to\infty}\mu_n(G)\geq \mu(G). $$
Proof. The cases \(G=\varnothing\) and \(G=\mathbb{R}^d\) are immediate. In every other case, use \(g_m\) above. Since \(g_m\leq 1_G\), for each \(n\) and \(m\),
$$ \mu_n(G)\geq \int g_m\,d\mu_n. $$

Fix \(m\). Weak convergence applies to the bounded continuous function \(g_m\), giving

$$ \liminf_{n\to\infty}\mu_n(G)\geq \int g_m\,d\mu. $$

The inequality holds for every \(m\). Because \(g_m\uparrow 1_G\), the monotone convergence theorem yields \(\int g_m\,d\mu\to\mu(G)\). Taking \(m\to\infty\) proves the lower bound.

\(\square\)

Worked Example: A Point Mass Approaching the Boundary

Worked Example: A Point Mass Approaching the Boundary

Let \(\mu_n=\delta_{1/n}\) on \(\mathbb{R}\), and let \(\mu=\delta_0\). For every bounded continuous function \(h\), continuity at zero gives \(\int h\,d\mu_n=h(1/n)\to h(0)=\int h\,d\mu\); hence \(\mu_n\Rightarrow\mu\).

Take the closed set \(F=\{0\}\). Then \(\mu_n(F)=0\) for every \(n\), while \(\mu(F)=1\), so the upper bound reads \(0\leq1\). The approximating function is \(f_m(x)=\max\{1-m|x|,0\}\). For fixed \(m\),

$$ \int f_m\,d\mu_n=\max\{1-m/n,0\}\longrightarrow 1 =\int f_m\,d\mu. $$

The event probability stays zero even though the limiting measure assigns probability one to \(F\). The continuous approximations first converge for each fixed \(m\); only after that limit do we let \(m\) increase and recover \(\mu(F)=1\).

Worked Example: Rare Mass Moving Far Away

Worked Example: Rare Mass Moving Far Away

For \(n\in\mathbb{N}\), let \(\mu_n=(1-1/n)\delta_0+(1/n)\delta_n\), and set \(\mu=\delta_0\). For any bounded continuous \(h\),

$$ \left|\int h\,d\mu_n-\int h\,d\mu\right| =\frac{1}{n}|h(n)-h(0)| \leq \frac{2\|h\|_\infty}{n}\longrightarrow 0. $$

Thus \(\mu_n\Rightarrow\mu\), even though a small amount of probability is placed at a point moving arbitrarily far away. For the closed set \(F=\{0\}\), \(\mu_n(F)=1-1/n\to1=\mu(F)\), consistent with the upper bound. For the open set \(G=\mathbb{R}\setminus\{0\}\), \(\mu_n(G)=1/n\to0=\mu(G)\), consistent with the lower bound.

The one-sided bounds do not rule out mass at locations depending on \(n\). Here the weight of that mass tends to zero, which is exactly why every bounded continuous test has a vanishing contribution from it.

Worked Example: Uniform Measures Near a Limit Point

Worked Example: Uniform Measures Near a Limit Point

Let \(\mu_n\) be the uniform probability measure on \([1/n,2/n]\), and let \(\mu=\delta_0\). For bounded continuous \(h\),

$$ \left|\int h\,d\mu_n-h(0)\right| \leq \sup_{x\in[1/n,2/n]}|h(x)-h(0)| \longrightarrow 0, $$

because every point in that interval approaches zero and \(h\) is continuous at zero. Hence \(\mu_n\Rightarrow\delta_0\). For \(F=\{0\}\), every \(\mu_n(F)=0\), while \(\mu(F)=1\). The closed-set upper bound holds but is strict.

For a fixed \(m\), when \(n>2m\), the function \(f_m(x)=\max\{1-m|x|,0\}\) equals \(1-mx\) throughout \([1/n,2/n]\). Its average is therefore

$$ \int f_m\,d\mu_n =1-m\left(\frac{1/n+2/n}{2}\right) =1-\frac{3m}{2n} \longrightarrow 1 =\int f_m\,d\mu. $$

This calculation again shows why the order matters: keep the approximation index fixed while using weak convergence, then send that index to infinity.

Why the Bounds Are One-Sided

The closed-set result is an upper bound because its continuous approximations lie above the indicator. The open-set result is a lower bound because its approximations lie below the indicator. Reversing either approximation direction would not justify the stated inequality.

A common pitfall is to expect \(\mu_n(A)\to\mu(A)\) for every set \(A\). Weak convergence does not guarantee this: the point-mass example has \(\mu_n(\{0\})=0\) for all \(n\) but \(\mu(\{0\})=1\). The Portmanteau bounds control closed and open sets in the appropriate direction without asserting equality. Convergence of probabilities requires additional conditions on the set; that question is treated through continuity sets.

The proof also depends on taking limits in the right order. For each fixed approximation index, weak convergence handles a bounded continuous function. The pointwise approximation is then used in a second limit under the fixed measure \(\mu\). One cannot replace these two steps with an unsupported claim that the indicators themselves are valid weak-convergence tests.

Check Your Understanding

Use the direction of approximation and the order of limits to answer the following questions.

  1. Why is \(d(x,A)\) continuous for every nonempty set \(A\subseteq\mathbb{R}^d\)?
  2. For a closed set \(F\), why does \(f_m(x)=\max\{1-md(x,F),0\}\) converge pointwise to \(1_F(x)\)?
  3. Which inequality between \(1_F\) and \(f_m\) leads to the closed-set upper bound?
  4. Why do we first fix \(m\) when applying weak convergence, rather than apply it directly to \(1_F\)?
  5. What do the examples show about the possibility that \(\mu_n(A)\) converges to \(\mu(A)\) for every set \(A\)?

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