Proof of the Dominated Convergence Theorem

From Fatou’s Lemma to Dominated Convergence

Pointwise convergence alone does not justify passing a limit through a Lebesgue integral. The Dominated Convergence Theorem supplies a sufficient condition: all functions in the sequence must be bounded in absolute value by one fixed integrable function. This tutorial proves the theorem using Fatou’s lemma, whose key role is to convert nonnegativity into an inequality for integrals.

Let \((X,\mathcal{M},\mu)\) be any measure space; it need not be finite or sigma-finite, and it need not be complete. A real-valued measurable function \(u\) is integrable when \(\int |u|\,d\mu<\infty\). We use Fatou’s lemma in the following form: if \(u_n:X\to[0,\infty]\) are measurable, then \(\int\liminf_n u_n\,d\mu\le\liminf_n\int u_n\,d\mu\). This is the standard Fatou lemma established in the earlier material on the Lebesgue integral.

The hypotheses are almost-everywhere statements, not necessarily pointwise statements everywhere. Since there are only countably many \(f_n\), we may take the union of the null sets on which domination fails, together with the null set on which convergence fails. This is still a measurable null set \(N\). Redefine every \(f_n\) and \(f\) to be zero on \(N\). These changes preserve measurability and integrals, and now the domination and convergence statements hold everywhere. We may therefore prove the theorem in this modified situation. Also, integrability of \(g\) implies \(g<\infty\) a.e., and \(|f|\le g\) follows by taking limits.

Proof of the Theorem

Proof. First, \(|f_n|\le g\) gives \(\int|f_n|\,d\mu\le\int g\,d\mu<\infty\), so every \(f_n\) is integrable. Taking the pointwise limit in \(|f_n|\le g\) gives \(|f|\le g\), so \(f\) is integrable as well. In particular, all signed integrals below are finite.

Because \(|f_n|\le g\), both \(g+f_n\) and \(g-f_n\) are nonnegative measurable functions. Their pointwise limits are \(g+f\) and \(g-f\), respectively. Applying Fatou’s lemma to \(g+f_n\) gives

The equality uses linearity of the integral and the fact that \(\int g\) is a finite constant. Subtracting \(\int g\,d\mu\) yields \(\int f\,d\mu\le\liminf_n\int f_n\,d\mu\). Applying Fatou’s lemma instead to the nonnegative functions \(g-f_n\) gives

Consequently, \(\limsup_n\int f_n\,d\mu\le\int f\,d\mu\). The sequence of integrals is bounded, since \(|\int f_n\,d\mu|\le\int g\,d\mu\). Its liminf is at least \(\int f\,d\mu\), and its limsup is at most \(\int f\,d\mu\); hence both equal \(\int f\,d\mu\), which proves convergence. More explicitly, equality of the liminf and limsup to this finite number means that for every \(\varepsilon>0\), all sufficiently large \(n\) satisfy \(\left|\int f_n\,d\mu-\int f\,d\mu\right|<\varepsilon\). \(\square\)

Proof. After the null-set modification described above, \(|f_n-f|\le|f_n|+|f|\le2g\) everywhere. Thus \(h_n=2g-|f_n-f|\) is nonnegative and measurable, and \(h_n\to2g\) pointwise because \(f_n\to f\). Fatou’s lemma gives

The last equality is valid because \(2\int g\) is finite and \(\int|f_n-f|\) is a bounded nonnegative sequence. It follows that the limsup of these nonnegative integrals is at most zero. Therefore they tend to zero: for every \(\varepsilon>0\), eventually \(0\le\int|f_n-f|\,d\mu<\varepsilon\). \(\square\)

For complex-valued measurable functions, the same convergence-of-integrals conclusion follows by applying the real theorem to the real and imaginary parts: \(|\operatorname{Re} f_n|\le g\), \(|\operatorname{Im} f_n|\le g\), and both parts converge a.e. The complex integral is defined by its real and imaginary parts. The \(L^1\) conclusion follows directly from the proof above, which uses only \(|f_n-f|\le|f_n|+|f|\).

Worked Applications

Why the Integrable Bound Matters

Consider \(f_n=n\chi_{(0,1/n)}\) on \((\mathbb{R},\mathcal{L},m)\). For each fixed \(x>0\), \(f_n(x)=0\) once \(n>1/x\), and for \(x\le0\) it is always zero. Thus \(f_n\to0\) pointwise. Nevertheless,

for every \(n\), whereas the integral of the pointwise limit is zero. The missing condition is a common integrable dominating function. Indeed, if a measurable \(g\ge0\) dominated every \(f_n\) a.e., then for each integer \(k\ge1\), \(g\ge k\) a.e. on \((1/(k+1),1/k)\), because \(k\chi_{(0,1/k)}\) is among the functions in the sequence. These intervals are disjoint and have measure \(1/(k(k+1))\). Consequently, for each \(K\ge1\),

The right side is unbounded as \(K\to\infty\), so no such \(g\) can be integrable. This pinpoints the gap: pointwise convergence describes the values at each fixed point, but without an integrable bound it does not control the total integral uniformly.