The Monotone Convergence Theorem

Why Increasing Limits Are Special

Pointwise convergence alone does not guarantee that the integrals converge to the integral of the pointwise limit. An increasing sequence of nonnegative functions has an additional structure: every function lies below the next one, so its integral cannot decrease. The Monotone Convergence Theorem identifies the limit of these integrals exactly, even when that limit is infinite.

Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; no assumption of finiteness, sigma-finiteness, or completeness is needed. Functions in the theorem are measurable and take values in \([0,\infty]\). The inequalities and convergence in its hypotheses hold at every \(x\in X\), and its conclusion is an equality of extended nonnegative numbers. Recall the supremum definition of the nonnegative integral:

A simple function here takes finitely many finite nonnegative values. We use two facts about this integral established earlier: it is monotone, so \(0\le g\le h\) implies \(\int g\,d\mu\le\int h\,d\mu\); and if \(\varphi=\sum_{j=1}^k a_j\chi_{A_j}\), with disjoint measurable \(A_j\) and \(a_j\ge0\), then \(\int\varphi\,d\mu=\sum_{j=1}^k a_j\mu(A_j)\). We also use continuity from below for measures: if \(B_n\) increases to \(B\), then \(\mu(B_n)\to\mu(B)\), with \(+\infty\) allowed.

A Simple-Function Approximation Lemma

Proof. Write \(\varphi=\sum_{j=1}^k a_j\chi_{A_j}\) with disjoint measurable \(A_j\) and finite coefficients \(a_j\ge0\). Then \(\int_{E_n}\varphi\,d\mu=\sum_{j=1}^k a_j\mu(E_n\cap A_j)\). For each \(j\), the sets \(E_n\cap A_j\) increase to \(A_j\), so continuity from below gives \(\mu(E_n\cap A_j)\to\mu(A_j)\). The sum has only finitely many nonnegative terms, so its limit is \(\sum_{j=1}^k a_j\mu(A_j)=\int_X\varphi\,d\mu\). This remains true if a term, or the sum, tends to \(+\infty\); zero coefficients contribute zero. \(\square\)

The Monotone Convergence Theorem

Proof. First, \(f\) is measurable: for every real \(a\), \(\{f>a\}=\bigcup_{n=1}^{\infty}\{f_n>a\}\), a measurable set. By monotonicity of the integral, the numbers \(\int f_n\,d\mu\) form a nondecreasing sequence. Write \(L=\lim_n\int f_n\,d\mu\), allowing \(L=+\infty\). Since \(f_n\le f\), each \(\int f_n\,d\mu\le\int f\,d\mu\), hence \(L\le\int f\,d\mu\).

For the reverse inequality, fix any nonnegative simple function \(\varphi\le f\), and fix \(t\) with \(0<t<1\). Define \(E_n=\{x:f_n(x)\ge t\varphi(x)\}\). These sets are measurable and increase with \(n\). We claim their union is \(X\). At a point where \(\varphi(x)=0\), the inequality defining \(E_n\) holds for every \(n\). At a point where \(\varphi(x)>0\), the value \(\varphi(x)\) is finite and \(t\varphi(x)<\varphi(x)\le f(x)\). Thus \(f_n(x)\to f(x)\) implies \(f_n(x)>t\varphi(x)\) for some \(n\). This argument also applies when \(f(x)=+\infty\). Therefore \(E_n\uparrow X\).

On \(E_n\), \(f_n\ge t\varphi\), and outside \(E_n\), \(t\varphi\chi_{E_n}=0\). Consequently, monotonicity and the scaling rule for the integral give

By the lemma, \(\int_{E_n}\varphi\,d\mu\to\int\varphi\,d\mu\). Taking limits in the displayed inequality yields \(L\ge t\int\varphi\,d\mu\). If \(\int\varphi\,d\mu\) is finite, letting \(t\) increase to \(1\) gives \(L\ge\int\varphi\,d\mu\). If \(\int\varphi\,d\mu=+\infty\), the right side \(t\int\varphi\,d\mu\) is \(+\infty\) for every \(t>0\), so the same inequality forces \(L=+\infty\). Thus in either case \(L\ge\int\varphi\,d\mu\). Taking the supremum over all nonnegative simple \(\varphi\le f\), as required by the definition of \(\int f\,d\mu\), gives \(L\ge\int f\,d\mu\). Combined with the opposite inequality already proved, this establishes the theorem. \(\square\)

The proof’s key step is the comparison with \(t\varphi\), where \(t<1\). The strict gap between \(t\varphi\) and \(\varphi\) ensures that each point is eventually included in some \(E_n\), even if \(f_n(x)\) approaches \(f(x)\) without ever equalling it. The supremum definition then transfers the comparison for simple functions to the full integral.

Worked Applications

A Common Pitfall: Pointwise Convergence Is Not Enough

Do not apply the theorem to a sequence that merely converges pointwise. On \(\mathbb{R}\) with Lebesgue measure, set \(g_n=n\chi_{(0,1/n)}\). For every fixed \(x>0\), eventually \(1/n\le x\), so \(g_n(x)=0\) from then on; at \(x\le0\), \(g_n(x)=0\) for every \(n\). Thus \(g_n\to0\) everywhere. But \(m((0,1/n))=1/n\), and hence \(\int g_n\,dm=n(1/n)=1\) for every \(n\), whereas the integral of the limit is \(0\).

The missing condition is monotonicity. For example, at \(x=3/4\), \(g_1(x)=1\) because \(3/4\in(0,1)\), while \(g_2(x)=0\) because \(3/4\notin(0,1/2)\). Thus \(g_1(x)>g_2(x)\), contradicting the required inequality \(g_1\le g_2\). The theorem supplies a precise condition under which the limit and integral agree; pointwise convergence on its own supplies no such guarantee.