From a One-Sided Bound to Convergence
Fatou’s lemma gives a one-sided inequality: the integral of a pointwise lower limit is at most the lower limit of the integrals. To prove that integrals actually converge when functions converge pointwise, we need more control than Fatou’s lemma alone provides. The key additional hypothesis is a single integrable function that bounds every member of the sequence.
Throughout, \((X,\mathcal A,\mu)\) is a measure space, and integrals are Lebesgue integrals. The functions in the theorem below are real-valued and measurable. An integrable dominating function prevents large values from accumulating in regions that move with the index, as happens in examples where pointwise convergence alone does not control the integrals.
The bound has two jobs. First, it ensures each \(f_n\) is integrable, because \(|f_n|\leq g\) almost everywhere and \(g\) has finite integral. Second, it passes to the pointwise limit: wherever \(f_n(x)\to f(x)\) and \(|f_n(x)|\leq g(x)\), continuity of absolute value gives \(|f(x)|\leq g(x)\). Thus the limit is integrable as well. These facts make the differences of integrals in the proof legitimate finite real numbers.
The Dominated Convergence Theorem
Proof. There is a measurable null set outside which both the pointwise convergence and all the domination inequalities hold. Indeed, the exceptional set for convergence is null, and the union over \(n\) of the null sets where \(|f_n|\leq g\) fails is still null. Redefine every \(f_n\) and \(f\) to be zero on this union. These changes do not affect their integrals, and now convergence and domination hold at every point. We prove the theorem for these modified functions.
For every \(n\), \(|f_n|\leq g\), so \(f_n\) is integrable. Taking the limit in this pointwise inequality gives \(|f|\leq g\), so \(f\) is integrable too. Write \(G=\int_X g\,d\mu\), which is finite, and \(I_n=\int_X f_n\,d\mu\). The functions \(g+f_n\) and \(g-f_n\) are nonnegative and measurable. They converge pointwise to \(g+f\) and \(g-f\), respectively. Fatou’s lemma applied to the first sequence gives \( \displaystyle \int_X(g+f)\,d\mu\leq\liminf_{n\to\infty}\int_X(g+f_n)\,d\mu. \) By linearity of the integral for integrable functions, this becomes \( \displaystyle G+\int_X f\,d\mu\leq G+\liminf_{n\to\infty} I_n, \) so \(\int_X f\,d\mu\leq\liminf_{n\to\infty} I_n\).
Apply Fatou’s lemma a second time, now to \(g-f_n\). It yields \( \displaystyle \int_X(g-f)\,d\mu\leq\liminf_{n\to\infty}\int_X(g-f_n)\,d\mu. \) Since \(I_n\) lies between \(-G\) and \(G\), its lower and upper limits are finite, and the right-hand side equals \(G-\limsup_{n\to\infty}I_n\). Therefore \( \displaystyle G-\int_X f\,d\mu\leq G-\limsup_{n\to\infty}I_n, \) which implies \(\limsup_{n\to\infty}I_n\leq\int_X f\,d\mu\). Combining the two bounds gives \( \displaystyle \limsup_{n\to\infty}I_n\leq\int_X f\,d\mu\leq\liminf_{n\to\infty}I_n. \) Because the lower limit of a real sequence cannot exceed its upper limit, both are equal to \(\int_X f\,d\mu\). Hence \(I_n\) converges to that integral. \(\square\)
The two applications of Fatou’s lemma are the central proof technique. Applying it only to \(g+f_n\) gives a lower bound on the limit integral. Applying it to \(g-f_n\) reverses the comparison and gives an upper bound. The common integrable function \(g\) makes both transformed sequences nonnegative while keeping all the integrals finite.
Worked Applications
Worked Example: Powers on a Closed Interval
On \([0,1]\) with Lebesgue measure, let \(f_n(x)=x^n\). For every \(x\in[0,1)\), \(x^n\to0\), while \(f_n(1)=1\) for all \(n\). Thus the pointwise limit is zero except at the single point \(1\). The function \(g(x)=1\) dominates every \(f_n\), since \(0\leq x^n\leq1\), and it is integrable on \([0,1]\). The limit function is zero almost everywhere, so its integral is zero. Directly, \( \displaystyle \int_0^1 x^n\,dx=\frac{1}{n+1}\longrightarrow0. \) The Dominated Convergence Theorem therefore gives the same conclusion as the calculation. The value of the limit at \(x=1\) does not change its integral, because a singleton has Lebesgue measure zero.
Worked Example: A Sequence on an Unbounded Interval
On \([0,\infty)\) with Lebesgue measure, define \( \displaystyle f_n(x)=e^{-x}\bigl(1+\sin(x/n)\bigr). \) For each fixed \(x\), \(\sin(x/n)\to0\), so \(f_n(x)\to e^{-x}\). Since \(0\leq1+\sin(x/n)\leq2\), the functions satisfy \(|f_n(x)|\leq2e^{-x}\), and \(2e^{-x}\) is integrable with integral \(2\). The theorem gives \( \displaystyle \int_0^\infty f_n(x)\,dx\longrightarrow\int_0^\infty e^{-x}\,dx=1. \) The conclusion can also be checked by bounding the error. Using \(|\sin u|\leq|u|\), \( \displaystyle \left|\int_0^\infty f_n(x)\,dx-1\right| \leq\int_0^\infty e^{-x}|\sin(x/n)|\,dx \leq\frac1n\int_0^\infty xe^{-x}\,dx =\frac1n. \) The last integral equals \(1\), by integration by parts. This example illustrates that a common integrable bound can control the tails even when the domain has infinite measure.
Worked Example: Convergence of a Sequence of Series Terms
Let \(X=\mathbb N\) with counting measure, and set \( \displaystyle f_n(k)=\begin{cases}(-1)^k/k^2,&k\leq n,\\0,&k>n.\end{cases} \) For each fixed \(k\), once \(n\geq k\), \(f_n(k)=(-1)^k/k^2\). Hence the pointwise limit is \(f(k)=(-1)^k/k^2\). Take \(g(k)=1/k^2\). It dominates every \(f_n\), and it is summable: for \(k\geq2\), \( \displaystyle \frac1{k^2}\leq\frac1{k(k-1)}=\frac1{k-1}-\frac1k. \) The latter terms telescope, so \(\sum_{k=1}^\infty1/k^2\leq2\). The Dominated Convergence Theorem now gives \( \displaystyle \sum_{k=1}^n\frac{(-1)^k}{k^2}\longrightarrow\sum_{k=1}^\infty\frac{(-1)^k}{k^2}. \) Here integration with respect to counting measure is summation. The theorem packages convergence of these partial sums as an instance of convergence of integrals.
Worked Example: Pointwise Convergence Without an Integrable Bound
On \(\mathbb N\) with counting measure, define \(h_n(k)=n\) if \(k=n\), and \(h_n(k)=0\) otherwise. For each fixed \(k\), \(h_n(k)=0\) whenever \(n>k\), so \(h_n(k)\to0\). Nevertheless, \( \displaystyle \int_{\mathbb N}h_n\,d\mu=\sum_{k=1}^\infty h_n(k)=n, \) which does not converge to the integral of the pointwise limit, namely zero. There cannot be an integrable function \(g\) dominating all the \(h_n\): such a bound would require \(g(n)\geq n\) for every \(n\), and therefore \( \displaystyle \int_{\mathbb N}g\,d\mu=\sum_{k=1}^\infty g(k)\geq\sum_{k=1}^\infty k=+\infty. \) This example pinpoints the missing hypothesis. Pointwise convergence by itself cannot prevent the mass of a function from moving to new locations.
Convergence of the Absolute Error
The theorem also yields convergence in integral of the absolute difference. This stronger-looking conclusion is a useful consequence, not an extra assumption. It says that the total integrated error between \(f_n\) and its limit becomes small.
Proof. We already know that \(|f|\leq g\) almost everywhere. Therefore \( \displaystyle |f_n-f|\leq|f_n|+|f|\leq2g \) almost everywhere. The functions \(|f_n-f|\) are measurable, converge pointwise almost everywhere to zero, and are dominated by the integrable function \(2g\). Applying the Dominated Convergence Theorem to this sequence gives \( \displaystyle \int_X|f_n-f|\,d\mu\longrightarrow\int_X0\,d\mu=0. \) This proves the corollary. \(\square\)
Choosing and Checking the Bound
A common mistake is to find a bound that works separately for each \(n\) and assume that this is enough. The theorem requires one function \(g\) that dominates the entire sequence and has finite integral. Another mistake is to use a constant bound on a space of infinite measure: a nonzero constant need not be integrable there. In that setting, the example on \([0,\infty)\) uses a decaying bound instead.
The theorem does not require the sequence \(f_n\) to be monotone, nor does it require pointwise convergence at every point. Convergence almost everywhere is sufficient because changing functions on a null set does not affect their integrals. The essential checks are measurability, almost-everywhere convergence, and a common integrable bound. When these hold, Fatou’s lemma applied to both signs converts pointwise convergence into convergence of the integrals.
Check that \(f_n(x)\to f(x)\) outside a set of measure zero.
Verify \(|f_n|\leq g\) almost everywhere for every \(n\), with \(\int_Xg\,d\mu<\infty\).
The bound implies that each \(f_n\) and the pointwise limit \(f\) are integrable.
Conclude that the integrals converge; apply the theorem to \(|f_n-f|\) when integrated absolute error is needed.
Check Your Understanding
Use the theorem and its proof to answer these questions.
- Why does the pointwise limit \(f\) satisfy \(|f|\leq g\) wherever convergence and domination both hold?
- Why are both \(g+f_n\) and \(g-f_n\) nonnegative in the proof?
- Which inequality follows from applying Fatou’s lemma to \(g+f_n\), and which follows from applying it to \(g-f_n\)?
- In the counting-measure counterexample, why can no integrable function dominate every \(h_n\)?
- How does the Dominated Convergence Theorem imply that \(\int_X|f_n-f|\,d\mu\) tends to zero?