From Probability Bounds to Integral Bounds
Weak convergence controls integrals of bounded continuous functions. Portmanteau also supplies inequalities for probabilities of open and closed sets. The useful next step is to pass from bounds on sets to bounds on integrals of functions whose level sets have the appropriate openness.
For a lower-semicontinuous function, the sets where the function exceeds a threshold are open. The open-set form of Portmanteau therefore controls the probability of each such level set. Integrating those probability bounds over all thresholds yields a bound for the expected value of the function.
This is the open-set inequality from Portmanteau. The argument below shows how it extends from indicator functions of open sets to general bounded semicontinuous functions.
Semicontinuity and Level Sets
These conditions encode one-sided control near each point. Lower semicontinuity prevents nearby function values from dropping abruptly below the value at a point; upper semicontinuity prevents them from jumping abruptly above it. A continuous function has both properties, but semicontinuous functions may have discontinuities.
Every lower-semicontinuous function is Borel measurable: its strict superlevel sets are open, and these sets determine its measurability. Consequently its integral against a probability measure is well-defined when it is bounded. The corresponding statement for an upper-semicontinuous function follows by considering its negative.
Worked Example: An Open-Set Indicator
Worked Example: An Open-Set Indicator
Let \(Z\) have the standard normal distribution, let \(X_n=Z/\sqrt{n}\), and let \(\mu_n\) be the law of \(X_n\). Since \(X_n\to 0\) almost surely, bounded convergence applied to every bounded continuous test function gives \(\mu_n\Rightarrow\delta_0\), where \(\delta_0\) is point mass at zero.
Take \(f(x)=\mathbf{1}_{(0,\infty)}(x)\). This function is lower-semicontinuous: for \(0\leq a<1\), its strict superlevel set is \((0,\infty)\); for \(a<0\), it is \(\mathbb{R}\); and for \(a\geq1\), it is empty. All these sets are open.
Symmetry and the continuous normal distribution give \(\int f\,d\mu_n=\mathbb{P}(X_n>0)=1/2\) for every \(n\). But \(\int f\,d\delta_0=f(0)=0\). Thus the lower bound reads \(1/2\geq0\), as required. The inequality need not be an equality when the function is discontinuous at a point carrying mass under the limit law.
The Layer-Cake Identity
The main proof uses a representation of a nonnegative bounded function as the accumulated lengths of its strict superlevel sets. If \(g\) is measurable and \(0\leq g\leq C\), then for every \(x\), $$ g(x)=\int_0^C \mathbf{1}_{\{g(x)>t\}}\,dt. $$ Indeed, the integrand is one precisely for thresholds \(t\) below \(g(x)\), an interval of length \(g(x)\); its value at the single endpoint does not affect the integral.
Tonelli’s theorem, applied to the nonnegative indicator on the product of the probability space and the threshold interval, then gives $$ \int g\,d\nu=\int_0^C \nu(\{x:g(x)>t\})\,dt $$ for any probability measure \(\nu\). This identity turns a function integral into an integral of probabilities. It is exactly suited to the open-set Portmanteau inequality when \(g\) is lower-semicontinuous.
Lower Bound for Bounded Lower-Semicontinuous Functions
Choose finite constants \(m\) and \(M\) such that \(m\leq f(x)\leq M\) for every \(x\), and set \(C=M-m\). If \(C=0\), then \(f\) is constant and both sides of the claimed inequality are equal. Suppose henceforth that \(C>0\), and put \(g=f-m\). Then \(0\leq g\leq C\), and \(g\) is lower-semicontinuous.
For every \(t\in[0,C]\), the set \(G_t=\{x:g(x)>t\}\) is open. Portmanteau therefore gives \(\liminf_{n\to\infty}\mu_n(G_t)\geq\mu(G_t)\) for every such \(t\). Apply the layer-cake identity to \(g\) under each measure. Fatou’s lemma for the nonnegative measurable functions \(t\mapsto\mu_n(G_t)\) yields $$ \begin{aligned} \liminf_{n\to\infty}\int g\,d\mu_n &=\liminf_{n\to\infty}\int_0^C\mu_n(G_t)\,dt\\ &\geq\int_0^C\liminf_{n\to\infty}\mu_n(G_t)\,dt\\ &\geq\int_0^C\mu(G_t)\,dt =\int g\,d\mu. \end{aligned} $$ The functions of \(t\) being integrated are measurable because each is a nonincreasing tail-probability function. Finally, since all the measures are probability measures, $$ \int g\,d\mu_n=\int f\,d\mu_n-m \quad\text{and}\quad \int g\,d\mu=\int f\,d\mu-m. $$ Adding \(m\) to the inequality proves the result.
\(\square\)
Upper Bound for Bounded Upper-Semicontinuous Functions
Upper semicontinuity of \(f\) means that \(-f\) is lower-semicontinuous: for every real \(a\), the set \(\{-f>a\}\) equals \(\{f<-a\}\), which is open. The function \(-f\) is also bounded. Applying the lower-bound theorem to \(-f\) gives $$ \liminf_{n\to\infty}\int(-f)\,d\mu_n\geq\int(-f)\,d\mu. $$ Since \(\int(-f)\,d\nu=-\int f\,d\nu\) for every probability measure \(\nu\), and \(\liminf(-a_n)=-\limsup a_n\) for a bounded real sequence \((a_n)\), this inequality is equivalent to $$ \limsup_{n\to\infty}\int f\,d\mu_n\leq\int f\,d\mu. $$
\(\square\)
Worked Example: A Closed-Set Indicator
Worked Example: A Closed-Set Indicator
Use again the measures \(\mu_n\) that are the laws of \(Z/\sqrt{n}\), which converge weakly to \(\delta_0\). Define \(h(x)=\mathbf{1}_{\{0\}}(x)\). For \(0\leq a<1\), the strict sublevel set \(\{x:h(x)<a\}\) is empty; for \(a=1\), it is \(\mathbb{R}\setminus\{0\}\); and for \(a>1\), it is all of \(\mathbb{R}\). For \(a<0\), it is empty as well. Each set is open, so \(h\) is upper-semicontinuous.
Because \(Z/\sqrt n\) has a continuous distribution, \(\int h\,d\mu_n=\mathbb{P}(Z/\sqrt n=0)=0\) for every \(n\). On the other hand, \(\int h\,d\delta_0=h(0)=1\). The upper bound is therefore \(0\leq1\). This example also displays why the inequality may be strict: the limiting measure assigns positive mass to the point where the function is discontinuous.
Worked Example: A Non-Indicator Function
Worked Example: A Non-Indicator Function
Let \(\mu_n=\delta_{1/n}\), so \(\mu_n\Rightarrow\delta_0\): for every bounded continuous \(\varphi\), continuity at zero gives \(\int\varphi\,d\mu_n=\varphi(1/n)\to\varphi(0)=\int\varphi\,d\delta_0\). Define the continuous clipped function \(c(x)=\min\{1,\max\{0,x\}\}\), and set $$ f(x)=\frac12c(x)+\frac12\mathbf{1}_{(0,\infty)}(x). $$ The clipped function is continuous and bounded between zero and one. The indicator is lower-semicontinuous, so their nonnegative sum is lower-semicontinuous; directly, the only jump is at zero, where \(f(0)=0\) is no greater than the right-hand limiting value \(1/2\). Thus \(f\) is bounded and lower-semicontinuous.
For each \(n\in\mathbb{N}\), \(c(1/n)=1/n\), so $$ \int f\,d\mu_n=f(1/n)=\frac{1}{2n}+\frac12\longrightarrow\frac12, \qquad \int f\,d\delta_0=f(0)=0. $$ The theorem gives \(\frac12\geq0\). The function \(q=1-f\) is bounded and upper-semicontinuous, and its integrals satisfy \(\int q\,d\mu_n=\frac12-\frac{1}{2n}\to\frac12\), while \(\int q\,d\delta_0=1\). Its upper bound is \(\frac12\leq1\).
Why the Direction of the Inequality Matters
The lower bound is not a claim that the integrals converge. It only says that a bounded lower-semicontinuous test function cannot have its limiting integral exceed the lower limit of the approximating integrals. The upper-semicontinuous statement is the reverse one. Example indicators show the directions plainly: open-set indicators are lower-semicontinuous, while closed-set indicators are upper-semicontinuous.
For a bounded continuous function, both semicontinuity conditions hold. Applying the two inequalities gives matching lower and upper limits, hence convergence of the integrals. In general, however, applying the wrong inequality to a semicontinuous function reverses the conclusion and can lead to an invalid argument.
The global boundedness assumption ensures that these integrals are finite and that the layer-cake argument uses a finite threshold interval after shifting the function. Unbounded functions require additional conditions to control their tails; semicontinuity alone does not provide that control.
Check Your Understanding
Answer each question by identifying the relevant level sets or inequality.
- Which strict level sets must be open for a function to be lower-semicontinuous?
- Why does the layer-cake representation involve the sets \(\{g>t\}\) rather than \(\{g\geq t\}\)?
- Where in the proof of the lower integral bound is the open-set form of Portmanteau used?
- Explain how applying the lower bound to \(-f\) gives the upper bound for an upper-semicontinuous \(f\).
- For \(\mu_n=\delta_{1/n}\) and \(\mu=\delta_0\), why is \(\mathbf{1}_{(0,\infty)}\) lower-semicontinuous but not continuous at zero?