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Weak Convergence on Euclidean Spaces · Tutorial 165 of 946

Integrals of Bounded Functions with Null Discontinuity Sets

Learn when weak convergence guarantees convergence of expectations beyond bounded continuous test functions, and how null discontinuity sets make the extension possible.

Intermediate 10 min read

What You'll Learn

  • State the integral convergence theorem for bounded Borel functions with null discontinuity sets.
  • Prove the theorem using level sets and convergence on continuity sets.
  • Relate the theorem for indicator functions to convergence of event probabilities.
  • Apply the result to piecewise-continuous functions with negligible boundaries.
  • Identify why the discontinuity set must be negligible under the limiting law.

Beyond Continuous Test Functions

Weak convergence is defined through convergence of integrals against bounded continuous functions. In applications, however, the function of interest may have jumps. Indicators of events are a basic example: they are usually discontinuous at the event boundary.

The key extension is that boundedness can compensate for discontinuities when those discontinuities occur only on a set that has probability zero under the limiting law. The values of a function on such a set do not affect its limiting integral, and the exceptional set is small enough to control the integrals of nearby laws.

Central principle: If probability measures converge weakly, then integrals also converge for bounded Borel functions whose discontinuity set has measure zero under the limiting probability measure.

Setup and a Continuity-Set Fact

Let \(\mu_n\) and \(\mu\) be probability measures on \(\mathbb{R}^d\). We write \(\mu_n\Rightarrow\mu\) when \(\int g\,d\mu_n\to\int g\,d\mu\) for every bounded continuous function \(g:\mathbb{R}^d\to\mathbb{R}\). For a function \(f\), write \(D_f\) for its set of discontinuity points.

We will use the continuity-set consequence of the Portmanteau theorem: if \(A\) is Borel and \(\mu(\partial A)=0\), then \(\mu_n(A)\to\mu(A)\). Indeed, \(A^\circ\subseteq A\subseteq\overline A\), while the Portmanteau bounds give \(\liminf_n\mu_n(A^\circ)\geq\mu(A^\circ)\) and \(\limsup_n\mu_n(\overline A)\leq\mu(\overline A)\). The null-boundary condition makes both outer quantities equal to \(\mu(A)\), so the claimed convergence follows by squeezing.

Definition: A Borel set \(A\subseteq\mathbb{R}^d\) is a continuity set for \(\mu\) if \(\mu(\partial A)=0\). Here \(\partial A=\overline A\setminus A^\circ\) is the boundary of \(A\).

The Integral Convergence Theorem

Theorem: Suppose \(\mu_n\Rightarrow\mu\) on \(\mathbb{R}^d\). If \(f:\mathbb{R}^d\to\mathbb{R}\) is bounded and Borel measurable, and \(\mu(D_f)=0\), then $$ \int f\,d\mu_n\longrightarrow\int f\,d\mu. $$

The proof reduces the integral of a bounded nonnegative function to probabilities of its level sets. Almost every level set is a continuity set for \(\mu\), which allows the continuity-set result to be used.

Proof: First choose \(B>0\) such that \(|f(x)|\leq B\) for every \(x\). Set \(h=f+B\), so \(0\leq h\leq 2B\). The functions \(f\) and \(h\) have the same discontinuity set.

For each \(t\in\mathbb{R}\), let \(A_t=\{x:h(x)>t\}\). This is Borel. If \(x\notin D_h\) and \(h(x)\ne t\), continuity at \(x\) gives a neighborhood on which \(h\) stays strictly on the same side of \(t\) as \(h(x)\). Thus \(x\notin\partial A_t\), and consequently

$$ \partial A_t\subseteq D_h\cup\{x:h(x)=t\}. $$

The first set on the right has \(\mu\)-measure zero. There are at most countably many values \(t\) for which \(\mu(h=t)>0\): for each positive integer \(k\), at most \(k\) disjoint level sets can have measure at least \(1/k\), and every level set of positive measure has measure at least \(1/k\) for some \(k\). It follows that \(\mu(\partial A_t)=0\) for all but countably many \(t\). For each such \(t\), the continuity-set result gives

$$ \mu_n(A_t)\longrightarrow\mu(A_t). $$

For a nonnegative measurable function bounded by \(2B\), the layer-cake identity gives its integral as the integral of the measures of its strict upper level sets. Applying this identity to \(h\), and then using dominated convergence in the variable \(t\), yields

$$ \int h\,d\mu_n =\int_0^{2B}\mu_n(A_t)\,dt \longrightarrow \int_0^{2B}\mu(A_t)\,dt =\int h\,d\mu. $$

The integrands in the middle are bounded by \(1\), and the exceptional set of levels is countable, hence has Lebesgue measure zero. Finally, because \(\mu_n\) and \(\mu\) are probability measures,

$$ \int f\,d\mu_n=\int h\,d\mu_n-B \longrightarrow \int h\,d\mu-B=\int f\,d\mu. $$

This proves the theorem. \(\square\)

Worked Example: An Indicator of a Half-Line

Worked Example: An Indicator of a Half-Line

Let \(Z\) have the standard normal distribution, and let \(\mu_n\) be the law of \(Z+1/n\), with \(\mu\) the law of \(Z\). First, \(\mu_n\Rightarrow\mu\): for every bounded continuous \(g\), \(g(Z+1/n)\to g(Z)\) pointwise, and bounded convergence gives \(\mathbb{E}[g(Z+1/n)]\to\mathbb{E}[g(Z)]\).

Take \(f(x)=\mathbf{1}_{(0,\infty)}(x)\). Its discontinuity set is \(\{0\}\), and \(\mu(\{0\})=0\) because the standard normal law has a density. The theorem therefore applies. Writing \(\Phi\) for the standard normal distribution function, we can also calculate the probabilities directly:

$$ \int f\,d\mu_n =\mathbb{P}(Z+1/n>0) =\mathbb{P}(Z>-1/n) =\Phi(1/n) \longrightarrow \Phi(0)=\frac12 =\mathbb{P}(Z>0) =\int f\,d\mu. $$

This is convergence of event probabilities obtained from weak convergence because the boundary point of the event has zero limiting probability.

Worked Example: A Bounded Function with a Jump

Worked Example: A Bounded Function with a Jump

Continue with the same laws \(\mu_n\) and \(\mu\), and define

$$ f(x)= \begin{cases} 0,&x<0,\\ e^{-x},&x\geq 0. \end{cases} $$

This function is Borel measurable and bounded between \(0\) and \(1\). It is continuous except at \(0\), where its left limit is \(0\) and its value is \(1\). Since \(\mu(\{0\})=0\), the theorem gives convergence of its integrals.

The integrals can be evaluated explicitly. With \(\varphi\) the standard normal density and \(\Phi\) its distribution function, completing the square gives \(e^{-z}\varphi(z)=e^{1/2}\varphi(z+1)\). Therefore,

$$ \begin{aligned} \int f\,d\mu_n &=\int_{-1/n}^{\infty}e^{-(z+1/n)}\varphi(z)\,dz\\ &=e^{-1/n+1/2}\int_{1-1/n}^{\infty}\varphi(u)\,du\\ &=e^{-1/n+1/2}\Phi(1/n-1) \longrightarrow e^{1/2}\Phi(-1). \end{aligned} $$

Applying the same completion of the square to the limiting law gives \(\int f\,d\mu=\int_0^\infty e^{-z}\varphi(z)\,dz=e^{1/2}\Phi(-1)\), agreeing with the limit.

Worked Example: A Piecewise-Continuous Function

Worked Example: A Piecewise-Continuous Function

Again use the shifted normal laws, and let \(f(x)=x^2\) for \(|x|\leq 1\), while \(f(x)=0\) for \(|x|>1\). This is a bounded Borel function, continuous except at \(-1\) and \(1\). Both points have zero probability under \(\mu\), so its expectations converge.

The limiting expectation can be found by integration. Since \(\varphi'(x)=-x\varphi(x)\), an antiderivative of \(x^2\varphi(x)\) is \(-x\varphi(x)+\Phi(x)\). Hence

$$ \begin{aligned} \int f\,d\mu &=\int_{-1}^{1}x^2\varphi(x)\,dx\\ &=\big[-x\varphi(x)+\Phi(x)\big]_{-1}^{1}\\ &=2\Phi(1)-1-2\varphi(1). \end{aligned} $$

The theorem now gives the concrete conclusion

$$ \mathbb{E}\!\left[f(Z+1/n)\right] \longrightarrow 2\Phi(1)-1-2\varphi(1). $$

No continuity at the two endpoints is needed: what matters is that they carry no mass under the limiting normal law.

Indicators and Piecewise Continuity

For a Borel set \(A\), the indicator \(\mathbf{1}_A\) is discontinuous exactly at points of \(\partial A\). Applying the theorem to this indicator recovers the continuity-set result: if \(\mu(\partial A)=0\), then

$$ \mu_n(A)=\int\mathbf{1}_A\,d\mu_n \longrightarrow \int\mathbf{1}_A\,d\mu=\mu(A). $$

A useful sufficient condition for applying the theorem to a piecewise-continuous function is that it be bounded and continuous away from a set \(S\) with \(\mu(S)=0\). Then \(D_f\subseteq S\), so the theorem applies. For example, a bounded function on the real line that is continuous on each interval between finitely many breakpoints satisfies the condition whenever the limiting law assigns zero probability to every breakpoint.

The finiteness of the breakpoints is not essential. Countably many discontinuities are also permitted if their union has zero limiting probability. More generally, the theorem concerns the measure of the entire discontinuity set, not how complicated that set looks geometrically.

Why the Limiting Law Matters

The condition is specifically \(\mu(D_f)=0\), not \(\mu_n(D_f)=0\) for every \(n\). Weak convergence controls how mass behaves in the limit, and the exceptional set must be negligible for that limiting law. A jump point can be harmless when the limit assigns it probability zero, as in the examples above; it can be consequential when the limit places positive mass there.

Boundedness is also part of the theorem. It ensures the level-set integrals are controlled on a finite range and allows dominated convergence in the proof. For unbounded functions, weak convergence alone does not generally guarantee convergence of expectations; additional tail or integrability control is needed.

Practical test: To use this result, check three conditions: the laws converge weakly, the function is bounded and Borel measurable, and the limiting law gives probability zero to the function’s discontinuity set.

Check Your Understanding

Use the hypotheses of the theorem carefully when answering these questions.

  1. Why does \(\mu(\partial A)=0\) imply convergence of the integrals of \(\mathbf{1}_A\)?
  2. In the proof, why are there at most countably many levels \(t\) with \(\mu(h=t)>0\)?
  3. What property of \(h\) makes the layer-cake identity applicable on the interval from \(0\) to \(2B\)?
  4. Can a bounded function have a discontinuity and still satisfy the theorem’s hypothesis? Explain using the limiting measure.
  5. Which part of the argument would need additional control if \(f\) were unbounded?

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