Dominated Convergence on Infinite Measure Spaces

Domination on an Unbounded Domain

On a domain of infinite measure, a bound that is adequate on every bounded interval may fail to control the integral over the whole domain. For example, the constant function $1$ is bounded but has infinite integral on $[0,\infty)$. To pass a limit through an integral there, the essential task is to find one integrable function that bounds every member of the sequence, not merely to establish pointwise convergence or boundedness.

Throughout, let $m$ be Lebesgue measure on $[0,\infty)$, with the Lebesgue sigma-algebra. This measure space is sigma-finite and has infinite total measure. The functions in the examples are real-valued and measurable. We use the Dominated Convergence Theorem, whose hypotheses are stated here; its proof is the standard result established in the earlier material on the Dominated Convergence Theorem.

Theorem (Dominated Convergence): Let $(X,\mathcal{M},\mu)$ be a measure space, not necessarily of finite measure. Let $f_n:X\to\mathbb{R}$ be measurable, and suppose $f_n\to f$ $\mu$-a.e. for a measurable real-valued function $f$. If there is a measurable $g:X\to[0,\infty]$ with $\int g\,d\mu<\infty$ and $|f_n|\le g$ $\mu$-a.e. for every $n$, then $f$ and each $f_n$ are integrable and $\lim_{n\to\infty}\int f_n\,d\mu=\int f\,d\mu$. The measure $\mu(X)$ may be infinite.

The requirement is on the integral of the envelope $g$, not on the measure of the domain. Once a common integrable envelope is found, the theorem applies without restricting the domain to finite intervals or separately controlling tails. Two elementary envelope calculations will be useful.

Two Integrable Envelopes

Proposition: On $([0,\infty),\mathcal{L},m)$, for $a>0$ and an integer $k\ge0$, the function $x^k e^{-ax}$ is integrable and $$ \int_0^\infty x^k e^{-ax}\,dm(x)=\frac{k!}{a^{k+1}}. $$

Proof. First, $x^k e^{-ax}\to0$ as $x\to\infty$. Indeed, $e^{ax}\ge (ax)^{k+1}/(k+1)!$ for $x>0$, by the nonnegative-term power series for the exponential. Thus $0\le x^k e^{-ax}\le (k+1)!/(a^{k+1}x)$, and the right side tends to zero. For $k=0$, direct integration gives $\int_0^R e^{-ax}\,dx=(1-e^{-aR})/a$, whose limit is $1/a$. For $k\ge1$, integration by parts on $[0,R]$ gives $$ \int_0^R x^k e^{-ax}\,dx=-\frac{R^k e^{-aR}}{a}+\frac{k}{a}\int_0^R x^{k-1}e^{-ax}\,dx. $$ The boundary term tends to zero by the bound just proved. Induction on $k$, taking $R\to\infty$, now gives $\frac{k}{a}\frac{(k-1)!}{a^k}=\frac{k!}{a^{k+1}}$. The recurrence also shows each improper integral is finite. $\square$

Proposition: On $([0,\infty),\mathcal{L},m)$, if $p>1$, then $(1+x)^{-p}$ is integrable and $$ \int_0^\infty (1+x)^{-p}\,dm(x)=\frac{1}{p-1}. $$

Proof. For $R>0$, an antiderivative gives $$ \int_0^R(1+x)^{-p}\,dx=\frac{1-(1+R)^{1-p}}{p-1}. $$ Since $p>1$, $(1+R)^{1-p}\to0$ as $R\to\infty$. Taking this limit proves the formula and finiteness. $\square$

These formulas provide convenient envelopes: polynomial factors can be absorbed by exponential decay, while a rational tail comparable to $x^{-p}$ is integrable when $p>1$. The examples below use sequences with pointwise limits on the entire half-line, so no exceptional null set is needed.

Worked Limits on the Half-Line

Worked Example: A Family of Exponentials

For $n\ge1$, define $f_n(x)=e^{-(1+1/n)x}$ on $[0,\infty)$. Find its pointwise limit and the limit of its integrals.

For each fixed $x\ge0$, $1+1/n\to1$, so $f_n(x)\to f(x)=e^{-x}$. All functions are measurable and nonnegative. Since $1+1/n\ge1$, the exponentials satisfy $0\le f_n(x)\le e^{-x}$ for every $x\ge0$ and every $n$. The envelope is integrable: the first proposition with $a=1$ and $k=0$ gives $\int_0^\infty e^{-x}\,dm(x)=1$. Dominated Convergence therefore applies, and direct integration computes each term:

\int_0^\infty f_n(x)\,dm(x) =\frac{1}{1+1/n} =\frac{n}{n+1} \longrightarrow 1 =\int_0^\infty e^{-x}\,dm(x).

Check: substituting $n=1$ into the formula gives $1/(1+1)=1/2$; substituting $n=2$ gives $1/(1+1/2)=2/3$. Both agree with $n/(n+1)$, and these values approach the integral $1$.

Worked Example: A Rational Family

For $n\ge1$, let $f_n(x)=\dfrac{1}{(1+x)(1+x+1/n)}$ on $[0,\infty)$. Show that the integrals converge to the integral of the pointwise limit.

For every fixed $x\ge0$, the denominator tends to $(1+x)^2$, so $f_n(x)\to f(x)=(1+x)^{-2}$. The functions are measurable and nonnegative. Because $1+x+1/n\ge1+x$, we have $f_n(x)\le(1+x)^{-2}$. The rational-envelope proposition with $p=2$ gives $\int_0^\infty(1+x)^{-2}\,dm(x)=1$. Thus the envelope is integrable and the Dominated Convergence Theorem yields $\int f_n\,dm\to1$.

We can also compute each integral exactly. Put $\delta=1/n$, which is positive. Partial fractions give $$ \frac{1}{(1+x)(1+x+\delta)} =\frac{1}{\delta}\left(\frac{1}{1+x}-\frac{1}{1+x+\delta}\right). $$ Integrating from $0$ to $R$ gives $$ \int_0^R f_n(x)\,dm(x) =\frac{1}{\delta}\bigl(\ln(1+R)-\ln(1+R+\delta)+\ln(1+\delta)\bigr). $$ As $R\to\infty$, the difference $\ln(1+R)-\ln(1+R+\delta)=-\ln(1+\delta/(1+R))$ tends to zero. Therefore $$ \int_0^\infty f_n(x)\,dm(x) =n\ln(1+1/n). $$ For $t>0$, the identity $\ln(1+t)=\int_0^t(1+s)^{-1}\,ds$ implies $t/(1+t)\le\ln(1+t)\le t$, since $1/(1+t)\le1/(1+s)\le1$ for $0\le s\le t$. Taking $t=1/n$ and multiplying by $n$ gives $\frac{n}{n+1}\le n\ln(1+1/n)\le1$; both bounds tend to $1$. This independently confirms the limit. Check: for $n=1$, the exact integral is $\ln 2$, and for $n=2$, it is $2\ln(3/2)$; the squeeze bounds place each below $1$ and force the sequence to converge to $1$.

Worked Example: A Polynomial Factor with Exponential Decay

For $n\ge1$, define $f_n(x)=x e^{-(1+1/n)x}$ on $[0,\infty)$. Compute the limit of the integrals and verify it using a common envelope.

At every $x\ge0$, $f_n(x)\to f(x)=xe^{-x}$. The functions are measurable and nonnegative, and $1+1/n\ge1$ gives $0\le f_n(x)\le xe^{-x}$. The envelope is integrable by the first proposition with $a=1$ and $k=1$, with integral $1!/1^2=1$. Dominated Convergence applies. For each $n$, use that same proposition with $a=1+1/n$ and $k=1$:

\int_0^\infty x e^{-(1+1/n)x}\,dm(x) =\frac{1}{(1+1/n)^2} =\frac{n^2}{(n+1)^2} \longrightarrow 1 =\int_0^\infty xe^{-x}\,dm(x).

Check: at $n=1$, the formula gives $1/(1+1)^2=1/4$, which equals $1^2/(1+1)^2$. At $n=2$, it gives $1/(1+1/2)^2=4/9$, which equals $2^2/(2+1)^2$. The limit also agrees with the proposition’s value $1!/1^2=1$ for the limiting function.

What the Infinite Measure Changes

The theorem itself does not require a finite measure space. What changes is the usefulness of simple bounds: boundedness alone does not imply integrability. For example, on $([0,\infty),\mathcal{L},m)$, the constant function $1$ has integral $m([0,\infty))=\infty$. It cannot serve as the integrable envelope in the theorem, even if a particular sequence satisfies $|f_n|\le1$. The envelope must have finite integral over the full domain.

Common mistake. For the exponential sequence in the first example, one might write $|f_n(x)|\le1$ and then claim that Dominated Convergence applies with $g(x)=1$. The pointwise inequality is correct, but the application is not: $\int_0^\infty 1\,dm=\infty$, so the required integrable-envelope hypothesis fails. The correction is to use the sharper bound $|f_n(x)|\le e^{-x}$. Its integral is $1$, so it does meet the hypothesis, and the theorem applies.

Another necessary distinction is between pointwise convergence and convergence of integrals. For instance, on $([0,\infty),\mathcal{L},m)$, the real-valued measurable functions $h_n=\chi_{[n,n+1]}$ converge to zero at every fixed $x$: once $n>x$, $h_n(x)=0$. But $\int h_n\,dm=1$ for every $n$, whereas the integral of the pointwise limit is zero. These functions do not have a common integrable envelope: any nonnegative measurable $g$ with $g\ge h_n$ for all $n$ must satisfy $g\ge1$ on the union of the intervals $[n,n+1]$, which has infinite measure, and hence $\int g\,dm=\infty$. This illustrates why finding and checking the envelope is part of the argument, not an optional technicality.

For a sequence on an infinite-measure domain, a reliable verification is therefore to identify its pointwise limit, find a bound that works for every index on the whole domain, and compute or otherwise establish that the bound has finite integral. Only then does the theorem justify the limit of the integrals. In the examples, exponential decay supplied the envelope in two cases, while the rational denominator supplied an integrable power in the other.

Check Your Understanding

For each question, specify the measure space and check integrability of any proposed envelope before invoking the theorem.

On $[0,\infty)$ with Lebesgue measure, let $u_n(x)=e^{-(2+1/n)x}$. Find its pointwise limit and use an integrable envelope to compute $\lim_n\int_0^\infty u_n\,dm$.
For $v_n(x)=((1+x)(1+x+2/n))^{-1}$, find a common integrable rational envelope and determine the pointwise limit.
For $w_n(x)=x^2e^{-(1+1/n)x}$, identify an integrable envelope and compute the limit of the integrals using the exponential-polynomial formula.
Why does the bound $|f_n|\le C$, with a fixed finite constant $C$, not by itself provide an integrable envelope on $[0,\infty)$?
For $h_n=\chi_{[n,n+1]}$, verify pointwise convergence and compute every integral. Which Dominated Convergence hypothesis is unavailable?

Dominated Convergence on Infinite Measure Spaces

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