Applications of Dominated Convergence

From Convergence of Functions to Convergence of Integrals

The Dominated Convergence Theorem turns pointwise information about functions into convergence of their integrals, provided one integrable function controls the sequence. That makes it useful well beyond sequences written down in advance. A parameter-dependent family of functions gives a sequence whenever the parameter approaches a fixed value, and the theorem can then justify continuity or differentiation of the resulting integral.

We will use the Dominated Convergence Theorem and its $L^1$ consequence from earlier in this course. The focus here is how to check the hypotheses in applications. The essential questions are: are the functions measurable, do they converge almost everywhere, and is there a single integrable bound that works throughout the parameter values being considered?

Continuity of Parameter-Dependent Integrals

Let $f_t(x)$ denote a measurable function of $x$ for each parameter $t$ in an interval. The associated integral is $F(t)=\int_X f_t\,d\mu$, when $f_t$ is integrable. To prove continuity at $t_0$, consider any sequence of parameters $t_n\to t_0$. If $f_{t_n}\to f_{t_0}$ almost everywhere and the functions have a common integrable bound near $t_0$, the Dominated Convergence Theorem gives convergence of their integrals.

Theorem (Continuity of an Integral Under Domination): Let $J$ be an interval containing $t_0$, and suppose $f_t:X\to\mathbb{R}$ is measurable for every $t\in J$. Assume there is an integrable function $g:X\to[0,\infty]$ such that $|f_t|\leq g$ almost everywhere for every $t\in J$. If, whenever $t_n\in J$ and $t_n\to t_0$, we have $f_{t_n}\to f_{t_0}$ almost everywhere, then $F(t)=\int_X f_t\,d\mu$ is defined for every $t\in J$ and is continuous at $t_0$.

Proof. The bound $|f_t|\leq g$ almost everywhere and the integrability of $g$ imply that each $f_t$ is integrable. Now take any sequence $(t_n)$ in $J$ with $t_n\to t_0$. By hypothesis, $f_{t_n}\to f_{t_0}$ almost everywhere, and $|f_{t_n}|\leq g$ almost everywhere for every $n$. The Dominated Convergence Theorem therefore gives $$ \lim_{n\to\infty}\int_X f_{t_n}\,d\mu = \int_X f_{t_0}\,d\mu. $$ Thus $F(t_n)\to F(t_0)$ for every sequence $t_n\to t_0$ in $J$. The sequential criterion for continuity on a real interval implies that $F$ is continuous at $t_0$. This proves the theorem.

Worked Example: Integrating Over a Moving Interval

Let $X=[0,2]$ with Lebesgue measure, and for $t\in[0,2]$ define $f_t(x)=\mathbf{1}_{[0,t]}(x)$. Each $f_t$ is measurable and satisfies $|f_t(x)|\leq1$, where the constant function $1$ is integrable on $[0,2]$.

Fix $t_0\in(0,2)$, and let $t_n\to t_0$. For every $x\neq t_0$, membership of $x$ in $[0,t_n]$ eventually agrees with membership in $[0,t_0]$. Indeed, if $x<t_0$, then eventually $t_n>x$; if $x>t_0$, then eventually $t_n<x$. Thus $f_{t_n}(x)\to f_{t_0}(x)$ except possibly at $x=t_0$, a null set. The continuity theorem applies. Direct calculation identifies the integral: $$ F(t)=\int_0^2\mathbf{1}_{[0,t]}(x)\,dx=t. $$ So $F(t)$ is continuous, as the theorem predicts. The endpoints $t_0=0$ and $t_0=2$ can also be handled by the same argument with sequences restricted to $[0,2]$.

Differentiating Under the Integral Sign

Continuity requires control of the functions themselves. Differentiation requires control of difference quotients. A useful way to obtain that control is to bound the derivative of the integrand throughout a parameter interval. The ordinary Mean Value Theorem then bounds each difference quotient by the same integrable function.

Theorem (Differentiation Under the Integral Sign by Domination): Let $J$ be an open interval containing $t_0$. Suppose $f_t:X\to\mathbb{R}$ is measurable and integrable for every $t\in J$, and $f_{t_0}$ is integrable. Assume that outside a single null set, the function $t\mapsto f_t(x)$ is differentiable on $J$, its derivative at $t_0$ is a measurable function of $x$, and there is an integrable function $g:X\to[0,\infty]$ such that $$ \left|\frac{\partial}{\partial t}f_t(x)\right|\leq g(x) $$ for every $t\in J$. Then $F(t)=\int_X f_t\,d\mu$ is differentiable at $t_0$, and $$ F'(t_0)=\int_X\left.\frac{\partial}{\partial t}f_t\right|_{t=t_0}\,d\mu. $$

Proof. For $h\neq0$ sufficiently small that $t_0+h\in J$, define $$ q_h(x)=\frac{f_{t_0+h}(x)-f_{t_0}(x)}{h}. $$ For almost every $x$, the Mean Value Theorem applied to the real-valued function $t\mapsto f_t(x)$ gives a parameter $s$ between $t_0$ and $t_0+h$ such that $q_h(x)=\partial f_s(x)/\partial t$. Consequently, $|q_h(x)|\leq g(x)$ almost everywhere. For any sequence of nonzero $h_n\to0$, differentiability of $t\mapsto f_t(x)$ gives $$ q_{h_n}(x)\longrightarrow \left.\frac{\partial}{\partial t}f_t(x)\right|_{t=t_0} $$ almost everywhere. The difference quotients are measurable because they are formed from measurable functions. By the Dominated Convergence Theorem, $$ \lim_{n\to\infty}\int_X q_{h_n}\,d\mu = \int_X\left.\frac{\partial}{\partial t}f_t\right|_{t=t_0}\,d\mu. $$ Linearity of the integral gives $$ \int_X q_{h_n}\,d\mu = \frac{F(t_0+h_n)-F(t_0)}{h_n}. $$ The right-hand side therefore converges to the stated integral for every sequence of nonzero $h_n\to0$. By the sequential criterion for a real limit, the difference quotient for $F$ has that limit as $h\to0$. This proves differentiability and the formula.

The common null set in the hypotheses keeps the pointwise differentiability and derivative bound valid together for almost every $x$ throughout $J$. The integrable bound on the derivative controls the difference quotients; merely knowing that each $f_t$ is integrable would not provide that control.

Worked Example: Differentiating an Exponential Integral

On $[0,\infty)$ with Lebesgue measure, set $f_t(x)=e^{-tx}$ for $t>0$. Fix $t_0>0$ and take $J=(t_0/2,3t_0/2)$. For every $t\in J$, the function $f_t$ is integrable, and $$ \frac{\partial}{\partial t}f_t(x)=-xe^{-tx}. $$ For $t\in J$, we have $t>t_0/2$, and hence $$ \left|\frac{\partial}{\partial t}f_t(x)\right| =xe^{-tx} \leq xe^{-(t_0/2)x}. $$ The right-hand side is integrable, since $$ \int_0^\infty xe^{-(t_0/2)x}\,dx=\frac{4}{t_0^2}. $$ The differentiation theorem applies. In fact, $$ F(t)=\int_0^\infty e^{-tx}\,dx=\frac{1}{t}, $$ so $F'(t_0)=-1/t_0^2$. The integral of the derivative gives the same result: $$ \int_0^\infty -xe^{-t_0x}\,dx=-\frac{1}{t_0^2}. $$ The integrable bound is local in $t$: it works on $J$ around the chosen $t_0$, which is all the theorem requires.

Changing the Domain of Integration

Indicators let dominated convergence handle changing measurable sets. If the indicators of sets converge almost everywhere, multiplying them by a fixed integrable function preserves a useful integrable bound. This yields convergence of the integrals over those sets.

Corollary (Convergence of Integrals over Changing Sets): Let $f$ be an integrable real-valued measurable function, and let $A_n,A\in\mathcal{F}$. If $\mathbf{1}_{A_n}\to\mathbf{1}_A$ almost everywhere, then $$ \lim_{n\to\infty}\int_{A_n} f\,d\mu=\int_A f\,d\mu. $$

Proof. The functions $f\mathbf{1}_{A_n}$ are measurable and converge almost everywhere to $f\mathbf{1}_A$. Moreover, $|f\mathbf{1}_{A_n}|\leq|f|$, and $|f|$ is integrable. The Dominated Convergence Theorem gives convergence of their integrals. Using the definition of the integral over a measurable set, these integrals are exactly $\int_{A_n}f\,d\mu$ and $\int_Af\,d\mu$. This proves the corollary.

Worked Example: A Sequence Concentrating Near an Endpoint

On $[0,1]$, let $f_n(x)=x^n$. For every $x\in[0,1)$, $x^n\to0$, while at $x=1$, $x^n=1$. Thus $f_n\to0$ almost everywhere. Also $0\leq f_n\leq1$, and the constant function $1$ is integrable on $[0,1]$. The Dominated Convergence Theorem gives $$ \lim_{n\to\infty}\int_0^1x^n\,dx=0. $$ Indeed, direct integration verifies the conclusion: $$ \int_0^1x^n\,dx=\frac{1}{n+1}\longrightarrow0. $$ The functions remain equal to $1$ at the endpoint, but a single point does not affect the integral. This example also shows why convergence at every point is not needed: convergence almost everywhere is enough.

Why the Hypotheses Matter

The applications above rely on domination that is uniform over the sequence or parameter values used in the argument. For the continuity theorem, one bound $g$ must control $f_{t_n}$ as $t_n\to t_0$. For differentiation, one bound must control the difference quotients; the derivative bound supplies it through the Mean Value Theorem.

Without an integrable bound, pointwise convergence does not by itself allow the limit to pass through the integral. In the endpoint example, the bound $1$ is integrable because the interval has finite measure. On an infinite-measure space, a constant bound need not be integrable. Likewise, differentiability of each integrand alone is not enough to justify differentiating its integral: the difference quotients still need an integrable control.

In practice, the efficient order is to identify the parameter limit, establish almost-everywhere convergence, and then find a bound independent of the changing parameter. For differentiation, first bound the derivative on a neighborhood and use the Mean Value Theorem to transfer that bound to difference quotients. Once these checks are in place, the Dominated Convergence Theorem supplies the passage to the limit.

Key takeaway: Dominated convergence justifies continuity of parameterized integrals when the integrands converge under a common integrable bound. To differentiate under the integral sign, bound the derivatives locally so that the difference quotients are dominated as well.

Check Your Understanding

Use the hypotheses of dominated convergence to decide when a limit may pass through an integral.

For continuity of $F(t)=\int_X f_t\,d\mu$, what must happen to $f_{t_n}$ when $t_n\to t_0$, and what common bound is required?
In the differentiation theorem, which result bounds a difference quotient using a bound on the derivative?
Why must the derivative bound hold on a neighborhood of $t_0$, rather than only at $t_0$?
How does almost-everywhere convergence of $\mathbf{1}_{A_n}$ lead to convergence of integrals over $A_n$?
Why does a constant bound not automatically serve as an integrable dominating function on an infinite-measure space?