Proof Strategy for Dominated Convergence

A Localization Strategy

The Dominated Convergence Theorem is often proved using Fatou’s Lemma, as in the previous tutorial. A different proof strategy is useful when we want to see directly where the integral of the error becomes small. The idea is to divide the space into two parts: a region on which the dominating function is bounded and the measure is finite, and a remainder on which the integral of the dominating function is small.

On a measurable set $E$ with $\mu(E)<\infty$, Egorov’s Theorem, used here as a standard external result, states that if measurable finite-valued functions $u_n$ converge almost everywhere to $u$, then for every $\alpha>0$ there is a measurable $A\subseteq E$ with $\mu(A)<\alpha$ such that $u_n\to u$ uniformly on $E\setminus A$. Boundedness controls the error on that exceptional set. The integrable bound controls the error on the discarded remainder. This approach works even when the entire space has infinite measure.

Throughout, let $(X,\mathcal{F},\mu)$ be a measure space. The functions $f_n:X\to\mathbb{R}$, $f:X\to\mathbb{R}$, and $g:X\to[0,\infty]$ are measurable, and $g$ is integrable. We assume $f_n\to f$ almost everywhere and $|f_n|\leq g$ almost everywhere for every $n$. It is important that $f$ is assumed measurable: almost-everywhere convergence alone on an arbitrary, possibly incomplete measure space does not ensure that a proposed limit is measurable.

Building a Finite-Measure Core

An integrable function may be positive on a set of infinite measure, and it may be arbitrarily small over most of that set. Thus a useful finite-measure region cannot in general be obtained just by taking $\{g\leq M\}$. Instead, exclude both the region where $g$ is too small and the region where it is too large.

Lemma (Finite-Core Localization): Let $g:X\to[0,\infty]$ be measurable and integrable. For every $\varepsilon>0$, there are numbers $0<\delta<M<\infty$ such that, for $$ E=\{x\in X:\delta<g(x)\leq M\}, $$ we have $\mu(E)<\infty$ and $$ \int_{X\setminus E}g\,d\mu<\varepsilon. $$

Proof. The high-value tail can be made small: by the Vanishing Integral of the High-Value Tail result from earlier in the course, choose $M$ so that $$ \int_{\{g>M\}}g\,d\mu<\frac{\varepsilon}{2}. $$ For the low-value region, the functions $g\mathbf{1}_{\{g>1/k\}}$ increase pointwise to $g$ as $k\to\infty$. This includes points where $g=0$, since both sides are zero there; an integrable $g$ is finite almost everywhere. By the Monotone Convergence Theorem, their integrals increase to $\int_X g\,d\mu$. Since this integral is finite, choose $k$ large enough that $$ \int_{\{g\leq 1/k\}}g\,d\mu<\frac{\varepsilon}{2}. $$ Set $\delta=1/k$, increasing $M$ if necessary so that $M>\delta$. The sets $\{g\leq\delta\}$ and $\{g>M\}$ cover $X\setminus E$, so $$ \int_{X\setminus E}g\,d\mu \leq \int_{\{g\leq\delta\}}g\,d\mu+\int_{\{g>M\}}g\,d\mu <\varepsilon. $$ Finally, $\delta<g\leq M$ on $E$, and therefore $$ \delta\,\mu(E)\leq\int_E g\,d\mu\leq\int_Xg\,d\mu<\infty. $$ Because $\delta>0$, this proves $\mu(E)<\infty$.

The lower cutoff is essential. A bound $g\leq M$ alone does not imply that the region has finite measure: $g$ could be zero on a set of infinite measure. The cutoff $g>\delta$ gives the estimate $\delta\mu(E)\leq\int_Eg\,d\mu$, which guarantees finite measure.

The Error Estimate on the Core

The next result describes the main use of the core. It is a proof tool: once the tails are small, it remains to control the error on a finite-measure set where the dominating function is bounded.

Lemma (Finite-Core Error Estimate): Suppose $f_n\to f$ almost everywhere and $|f_n|\leq g$ almost everywhere for each $n$. Let $E=\{\delta<g\leq M\}$, where $0<\delta<M<\infty$ and $\mu(E)<\infty$. For every $\eta>0$, there is a measurable set $A\subseteq E$ with arbitrarily small measure such that, for all sufficiently large $n$, $$ \int_E|f_n-f|\,d\mu \leq 2M\mu(A)+\eta\mu(E). $$

Proof. First, $|f|\leq g$ almost everywhere: at every point outside the null set where convergence or domination fails, take the limit in $|f_n|\leq g$. Hence $|f_n-f|\leq 2g\leq 2M$ almost everywhere on $E$. By Egorov’s Theorem on the finite-measure set $E$, for any prescribed $\alpha>0$ there is a measurable $A\subseteq E$ with $\mu(A)<\alpha$ such that $f_n\to f$ uniformly on $E\setminus A$, apart from null sets if the convergence hypothesis holds only almost everywhere. Thus, for all sufficiently large $n$, $|f_n-f|<\eta$ almost everywhere on $E\setminus A$. Splitting the integral over $A$ and $E\setminus A$ gives $$ \int_E|f_n-f|\,d\mu \leq \int_A2M\,d\mu+\int_{E\setminus A}\eta\,d\mu \leq 2M\mu(A)+\eta\mu(E). $$ Since $\alpha$ can be chosen as small as desired, so can $\mu(A)$. If $\mu(E)=0$, the integral is already zero, and the claim holds as well. This proves the lemma.

Putting the Estimates Together

Here is the proof strategy for the $L^1$ conclusion of the Dominated Convergence Theorem. Combining this conclusion with the integral triangle inequality gives convergence of the signed integrals. We include the argument to show exactly how localization, Egorov’s Theorem, and domination fit together; it is an alternative to the Fatou’s Lemma proof in the previous tutorial.

Theorem (Dominated Convergence by Finite-Core Localization): Under the hypotheses stated above, $$ \lim_{n\to\infty}\int_X|f_n-f|\,d\mu=0. $$ Consequently, $\int_Xf_n\,d\mu\to\int_Xf\,d\mu$.

Proof. Given $\varepsilon>0$, apply the Finite-Core Localization Lemma to choose $E=\{\delta<g\leq M\}$ with $\mu(E)<\infty$ and $$ \int_{X\setminus E}g\,d\mu<\frac{\varepsilon}{4}. $$ On $X\setminus E$, the estimate $|f_n-f|\leq 2g$ holds almost everywhere. Therefore $$ \int_{X\setminus E}|f_n-f|\,d\mu \leq 2\int_{X\setminus E}g\,d\mu <\frac{\varepsilon}{2}. $$ On $E$, apply the Finite-Core Error Estimate with $\eta=\varepsilon/(4\mu(E))$ if $\mu(E)>0$. Choose its exceptional set $A$ so small that $2M\mu(A)<\varepsilon/4$. For all sufficiently large $n$, it follows that $$ \int_E|f_n-f|\,d\mu \leq 2M\mu(A)+\eta\mu(E) <\frac{\varepsilon}{2}. $$ If $\mu(E)=0$, this core integral is zero instead. Combining the core and remainder bounds gives $\int_X|f_n-f|\,d\mu<\varepsilon$ for all sufficiently large $n$. This proves $L^1$ convergence. The Dominated Convergence Theorem’s integrability conclusion follows from $|f_n|\leq g$ and $|f|\leq g$ almost everywhere. Finally, the integral triangle inequality gives $$ \left|\int_X f_n\,d\mu-\int_X f\,d\mu\right| \leq\int_X|f_n-f|\,d\mu\longrightarrow0. $$

Worked Applications

Worked Example: An Exponential Bound on an Infinite Interval

On $[0,\infty)$ with Lebesgue measure, let $g(x)=e^{-x}$ and $f_n(x)=e^{-x}\cos(x/n)$. These functions are measurable and satisfy $|f_n(x)|\leq e^{-x}=g(x)$. For each fixed $x$, $\cos(x/n)\to1$, so $f_n(x)\to f(x)=e^{-x}$. The limit is measurable.

Choose $0<\delta<1$ and $M\geq1$. The finite core is $$ E=\{x:\delta<e^{-x}\leq M\}=[0,\log(1/\delta)). $$ It has finite measure. The high-value tail is empty, and the integral on the low-value region is $$ \int_{\{e^{-x}\leq\delta\}}e^{-x}\,dx =\int_{\log(1/\delta)}^\infty e^{-x}\,dx =\delta. $$ Thus the discarded region has small $g$-integral when $\delta$ is small. On the core, $x$ lies in a bounded interval, so $x/n\to0$ uniformly there and $\cos(x/n)\to1$ uniformly. The integral error on the core therefore tends to zero, while the error outside it is at most $2\delta$. Taking $\delta$ arbitrarily small proves convergence of the integrals. Directly, $$ \int_0^\infty f_n(x)\,dx =\frac{1}{1+n^{-2}}\longrightarrow1 =\int_0^\infty e^{-x}\,dx. $$

Worked Example: Egorov’s Exceptional Set Near an Endpoint

On $[0,1]$, let $f_n(x)=e^{-nx}$, $f(x)=0$, and $g(x)=1$. At every $x>0$, $e^{-nx}\to0$; at $x=0$, convergence fails, but the singleton has measure zero. The functions are measurable and $|f_n|\leq g$, with $g$ integrable.

For $0<a<1$, remove $A=[0,a)$, whose measure is $a$. On the remaining interval $[a,1]$, the convergence is uniform because $$ \sup_{a\leq x\leq1}|f_n(x)-f(x)| =e^{-na}\longrightarrow0. $$ On the removed set, $|f_n-f|\leq1$, so its contribution to the error integral is at most $a$. On $[a,1]$, the contribution is at most $e^{-na}$, since that interval has measure at most $1$. Hence $$ \int_0^1|f_n-f|\,dx\leq a+e^{-na}. $$ First choose $a$ small, then choose $n$ large. This makes the integral arbitrarily small. The exact calculation confirms the result: $$ \int_0^1e^{-nx}\,dx=\frac{1-e^{-n}}{n}\longrightarrow0. $$ The estimate illustrates the order of choices in the proof: control the exceptional set first, then use uniform convergence on its complement.

Worked Example: Small Values on a Set of Infinite Measure

On $\mathbb{R}$ with Lebesgue measure, take $g(x)=1/(1+x^2)$ and $f_n(x)=g(x)\mathbf{1}_{[n,\infty)}(x)$. For every fixed $x$, $f_n(x)=0$ for all integers $n>x$, so $f_n\to0$ pointwise. Also, $|f_n|\leq g$, and $g$ is integrable.

The low-value region $\{g\leq\delta\}$ has infinite measure for every sufficiently small $\delta>0$. It would therefore be incorrect to estimate its integral by multiplying $\delta$ by the measure of that region. Instead, use the integral of $g$ itself. For $n\geq1$, $$ \int_{\mathbb{R}}f_n(x)\,dx =\int_n^\infty\frac{1}{1+x^2}\,dx =\frac{\pi}{2}-\arctan(n) \longrightarrow0. $$ In the localization proof, the contribution from the discarded low-value region is controlled by $\int_{\{g\leq\delta\}}g\,d\mu$, which tends to zero by the Monotone Convergence Theorem argument in the lemma. Its measure need not be finite.

What the Strategy Does—and Does Not—Require

The finite-core proof depends on two separate controls. Integrability makes the integral of $g$ small on the discarded low- and high-value regions. On the remaining region, the positive lower cutoff ensures finite measure, while the upper cutoff ensures boundedness. Egorov’s Theorem applies only after this finite-measure reduction.

A common error is to claim that bounded convergence applies on the whole space merely because $f_n$ and $g$ are bounded. Boundedness does not make an infinite measure space finite, and a small uniform bound need not have a finite integral there. Another error is to use Egorov’s Theorem directly on a space of infinite measure. The core construction supplies the missing finite-measure hypothesis.

Measurability is a separate requirement, not a consequence to be silently inferred from almost-everywhere convergence. In the Dominated Convergence Theorem, assume the limit $f$ is measurable, as well as each $f_n$. This ensures that $|f_n-f|$ is measurable and its integral is defined. A null exceptional set may be ignored in integral estimates, but on a noncomplete measure space it does not automatically make every function defined through that set measurable.

Key takeaway: First discard regions where the integrable bound has little total integral. On the remaining finite-measure, bounded core, use Egorov’s Theorem and control its small exceptional set. Keeping the limit’s measurability explicit makes the argument valid on arbitrary measure spaces.

Check Your Understanding

Use the finite-core localization strategy to answer these questions.

Why does the core $\{\delta<g\leq M\}$ have finite measure, while $\{g\leq M\}$ need not?
Which earlier result controls the integral of $g$ where $g$ is small?
Why is Egorov’s Theorem applied to the core rather than to the entire space?
How are the integral of the error on the exceptional set and the integral on its complement estimated?
Why should measurability of the pointwise limit be stated explicitly on a possibly incomplete measure space?