When Can a Limit Pass Through an Integral?
A sequence of functions may converge at almost every point while its integrals behave in an unexpected way. The key question is what prevents the functions from placing more and more mass in smaller regions as the sequence progresses. The Dominated Convergence Theorem answers this question: if all the functions are bounded in absolute value by one integrable function, then their integrals converge to the integral of their pointwise limit.
Throughout, \((X,\mathcal{F},\mu)\) is a measure space. The functions considered below are real-valued and measurable. We use Fatou’s Lemma, the Monotonicity of the Lebesgue Integral, and the absolute-value bound for integrable functions established earlier in this course. “Almost everywhere” allows the stated convergence or inequality to fail on a measurable set of measure zero.
The bound is one common integrable function \(g\), not a different bound for each \(n\). The conclusion concerns signed integrals, so positive and negative values are both allowed. The theorem also proves that the limit is integrable, even if this was not known in advance.
Proof Using Fatou’s Lemma
Proof. Each \(f_n\) is integrable because \(|f_n|\leq g\) almost everywhere and \(g\) is integrable. Indeed, the Monotonicity of the Lebesgue Integral gives \(\int_X|f_n|\,d\mu\leq\int_Xg\,d\mu<\infty\).
Since \(f_n(x)\to f(x)\) wherever the sequence converges and \(|f_n(x)|\leq g(x)\) there, taking limits gives \(|f(x)|\leq g(x)\) almost everywhere. Thus \(\int_X|f|\,d\mu\leq\int_Xg\,d\mu<\infty\), so \(f\) is integrable as well.
We may disregard a single measurable null set so that both the convergence and all the domination inequalities hold everywhere else: take the union of the exceptional sets for convergence and for the inequalities \(|f_n|\leq g\). This is a null set by countable subadditivity. Changing the functions on that set does not change their integrals, so we may apply Fatou’s Lemma to the following nonnegative functions.
Write \(G=\int_Xg\,d\mu\) and \(a_n=\int_Xf_n\,d\mu\). The functions \(g+f_n\) and \(g-f_n\) are nonnegative almost everywhere, and converge almost everywhere to \(g+f\) and \(g-f\), respectively. Fatou’s Lemma applied to the first sequence gives $$ G+\int_Xf\,d\mu = \int_X(g+f)\,d\mu \leq \liminf_{n\to\infty}\int_X(g+f_n)\,d\mu = G+\liminf_{n\to\infty}a_n. $$ All these integrals are finite: \(0\leq g\pm f_n\leq 2g\) almost everywhere. Cancelling the finite number \(G\) yields \(\int_Xf\,d\mu\leq\liminf_{n\to\infty}a_n\).
Applying Fatou’s Lemma to \(g-f_n\) instead gives $$ G-\int_Xf\,d\mu \leq G-\limsup_{n\to\infty}a_n. $$ Here we used \(\liminf_{n\to\infty}(G-a_n)=G-\limsup_{n\to\infty}a_n\). The sequence \((a_n)\) is bounded because the absolute-value bound for integrable functions gives \(|a_n|\leq\int_X|f_n|\,d\mu\leq G\). Rearranging the displayed inequality yields \(\limsup_{n\to\infty}a_n\leq\int_Xf\,d\mu\). Combining the two bounds, $$ \int_Xf\,d\mu \leq \liminf_{n\to\infty}a_n \leq \limsup_{n\to\infty}a_n \leq \int_Xf\,d\mu. $$ The liminf and limsup are therefore equal to \(\int_Xf\,d\mu\), proving that \(\int_Xf_n\,d\mu\to\int_Xf\,d\mu\). This proves the theorem.
Worked Applications
Worked Example: Powers on a Bounded Interval
On \([0,1]\) with Lebesgue measure, let \(f_n(x)=x^n\). For every \(x\in[0,1)\), \(x^n\to0\), while \(f_n(1)=1\). Thus \(f_n\to0\) almost everywhere; the only exception is the singleton \(\{1\}\), which has measure zero. Also \(0\leq x^n\leq1\), and the constant function \(g(x)=1\) is integrable on \([0,1]\). The Dominated Convergence Theorem applies, and direct integration verifies the conclusion: $$ \int_0^1x^n\,dx=\frac{1}{n+1}\longrightarrow0 = \int_0^1 0\,dx. $$ The exceptional endpoint does not affect the integral or the almost-everywhere convergence hypothesis.
Worked Example: An Integrable Bound on an Infinite Interval
On \([0,\infty)\) with Lebesgue measure, define \(f_n(x)=e^{-x}\mathbf{1}_{[0,n]}(x)\). For each fixed \(x\geq0\), \(x\leq n\) for all sufficiently large \(n\), so \(f_n(x)\to e^{-x}\). The common bound \(g(x)=e^{-x}\) is integrable, since $$ \int_0^\infty e^{-x}\,dx=1, \qquad |f_n(x)|\leq e^{-x}. $$ Consequently, the theorem gives convergence of the integrals. In this case they can also be calculated: $$ \int_0^\infty f_n(x)\,dx = \int_0^n e^{-x}\,dx = 1-e^{-n} \longrightarrow 1 = \int_0^\infty e^{-x}\,dx. $$ The measure space has infinite measure, but that causes no difficulty: it is the integrability of the bound, not finiteness of the measure of the whole space, that matters.
Worked Example: A Moving Spike Without an Integrable Bound
On \([0,1]\), consider \(f_n(x)=n\mathbf{1}_{[0,1/n]}(x)\). For every \(x>0\), \(f_n(x)=0\) for all sufficiently large \(n\), so \(f_n\to0\) almost everywhere. Nevertheless, $$ \int_0^1f_n(x)\,dx = n\cdot\frac{1}{n} = 1 $$ for every \(n\), rather than tending to \(\int_0^1 0\,dx=0\). There is no nonnegative integrable function that dominates every \(f_n\) almost everywhere. To see why, suppose such a function \(g\) existed. Outside a single null set (the union of the exceptional sets for the countably many inequalities), \(g(x)\geq f_n(x)\) for every \(n\). For \(x\in(0,1]\), choose \(n=\lfloor1/x\rfloor\), the greatest integer not exceeding \(1/x\). Then \(nx\leq1\), so \(x\leq1/n\), and \(g(x)\geq n=\lfloor1/x\rfloor\) outside that null set. For each integer \(k\geq1\), on \((1/(k+1),1/k]\) we have \(\lfloor1/x\rfloor\geq k\). Hence its integral over \((0,1]\) is at least $$ \sum_{k=1}^{\infty} k\left(\frac{1}{k}-\frac{1}{k+1}\right) = \sum_{k=1}^{\infty}\frac{1}{k+1} = \infty. $$ This contradicts the integrability of \(g\). The example shows exactly what domination rules out: a fixed amount of integral can concentrate in narrower regions while the pointwise limit is zero almost everywhere.
Convergence in the Integral of the Difference
The theorem gives more than convergence of the signed integrals. The functions themselves approach one another in the integral of their absolute difference.
Proof. We have \(f_n-f\to0\) almost everywhere. Also, \(|f_n-f|\leq|f_n|+|f|\leq2g\) almost everywhere, and \(2g\) is integrable. Apply the Dominated Convergence Theorem to the nonnegative functions \(|f_n-f|\), with dominating function \(2g\). It follows that $$ \lim_{n\to\infty}\int_X|f_n-f|\,d\mu = \int_X0\,d\mu = 0. $$
This conclusion is stronger than merely knowing that the signed integrals converge: it says the total absolute discrepancy becomes small. For example, the integral triangle inequality from the previous tutorial then gives $$ \left|\int_X f_n\,d\mu-\int_X f\,d\mu\right| = \left|\int_X(f_n-f)\,d\mu\right| \leq \int_X|f_n-f|\,d\mu \longrightarrow0. $$ Thus convergence in \(L^1\) also directly implies convergence of the integrals.
A Useful Special Case and a Common Pitfall
On a finite-measure space, a sequence bounded by one constant has an integrable dominating function. Specifically, if \(\mu(X)<\infty\) and \(|f_n|\leq M\) almost everywhere for every \(n\), where \(M\) is finite, then \(g=M\mathbf{1}_X\) satisfies $$ \int_Xg\,d\mu=M\mu(X)<\infty. $$ Therefore, almost-everywhere convergence of such a sequence implies convergence of its integrals. This is a useful finite-measure special case of the Dominated Convergence Theorem.
A frequent mistake is to assume that pointwise convergence alone permits interchanging a limit and an integral. The moving-spike example shows otherwise. Another mistake is to check only that each \(f_n\) is integrable: individual integrability does not provide a common integrable bound. The hypothesis concerns all terms at once, through one function \(g\). Conversely, the bound need only hold almost everywhere for each \(n\); countable subadditivity lets us gather the exceptional sets into one null set.
Check Your Understanding
Use the hypotheses and conclusions of the Dominated Convergence Theorem to answer these questions.
- What must be true of the functions that dominate the sequence, and why must the same bound work for every index?
- Why is the pointwise limit integrable under the theorem’s hypotheses?
- Which two nonnegative sequences are used in the Fatou’s Lemma proof to bound the liminf and limsup of the signed integrals?
- Why does the moving-spike sequence have integrals equal to \(1\), even though it converges to zero almost everywhere?
- What additional convergence conclusion follows for \(\int_X|f_n-f|\,d\mu\)?