Monotone Convergence Theorem
1 Introduction
The monotone convergence theorem is a selection of related theorems proving, under certain conditions, the convergence of monotone sequences.
In its simplest form, it says that a non-decreasing bounded-above sequence of real numbers
\[ a_1 \leq a_2 \leq \ldots \leq K, \]
converges to its smallest upper bound, or supremum \(K\). Likewise, a non-increasing bounded-below sequence of real numbers converges to its largest lower bound, its infimum.
In more advanced mathematics, the monotone convergence theorem states that for sequences of non-negative pointwise-increasing measurable functions
\[ 0 \leq f_1(x)\leq f_2(x) \le \ldots , \]
taking the integral and the supremum can be interchanged with the result being finite if either one is finite.
2 Real Sequences
In its simplest form the theorem concerns monotone sequences of real numbers, where it is a direct consequence of the completeness of \(\mathbb{R}\).
Every bounded monotone sequence of real numbers converges. In particular, a non-decreasing sequence \((a_n)\) that is bounded above converges to its supremum,
\[ \lim_{n\to\infty} a_n = \sup_n a_n, \]
and a non-increasing sequence that is bounded below converges to its infimum.
Proof. Let \((a_n)\) be non-decreasing with \(a_n \leq K\) for all \(n\). The set \(S = \{a_n : n \in \mathbb{N}\}\) is non-empty and bounded above, so by the completeness (least upper bound) axiom it has a supremum \(L = \sup S\). We claim \(a_n \to L\).
Fix \(\epsilon > 0\). Since \(L\) is the least upper bound, \(L - \epsilon\) is not an upper bound of \(S\), so there exists an index \(N\) with \(a_N > L - \epsilon\). Because the sequence is non-decreasing, \(a_n \geq a_N > L - \epsilon\) for every \(n \geq N\), while \(a_n \leq L\) for all \(n\) since \(L\) is an upper bound. Hence
\[ L - \epsilon < a_n \leq L \quad\Longrightarrow\quad |a_n - L| < \epsilon \qquad \text{for all } n \geq N, \]
so \(a_n \to L = \sup_n a_n\). The non-increasing case follows by applying this result to \((-a_n)\).
3 Measurable Functions
If \(\left\{f_n : X \to [0, \infty)\right\}\) is a sequence of measurable functions on a measurable set \(X\) (with measure \(\mu\)) such that \(f_n \to f\) pointwise almost everywhere and \(f_1\leq f_2 \leq \ldots,\) then
\[ \lim_{n\to \infty} \int _X f_n ~ d\mu = \int_X f ~ d\mu. \]
Proof. Since changing each \(f_n\) and \(f\) on a \(\mu\)-null set alters none of the integrals, we may assume the convergence \(f_n \to f\) and the ordering \(f_1 \leq f_2 \leq \cdots\) hold at every point of \(X\); in particular \(f = \sup_n f_n\) pointwise.
As \(f_n \leq f_{n+1}\), monotonicity of the integral makes \(\left(\int_X f_n \, d\mu\right)\) a non-decreasing sequence in \([0,\infty]\), so it converges to some \(\alpha \in [0,\infty]\). Since \(f_n \leq f\) for every \(n\), we have \(\int_X f_n \, d\mu \leq \int_X f \, d\mu\), and letting \(n \to \infty\) gives
\[ \alpha \leq \int_X f \, d\mu. \]
For the reverse inequality, fix a simple measurable function \(s\) with \(0 \leq s \leq f\) and a constant \(c \in (0,1)\), and set
\[ E_n = \left\{ x \in X : f_n(x) \geq c\, s(x) \right\}. \]
Each \(E_n\) is measurable, the ordering \(f_n \leq f_{n+1}\) gives \(E_n \subseteq E_{n+1}\), and \(\bigcup_n E_n = X\): if \(s(x) = 0\) then \(x \in E_1\), while if \(s(x) > 0\) then \(c\, s(x) < s(x) \leq f(x)\) and \(f_n(x) \uparrow f(x)\), so \(x \in E_n\) for all large \(n\). Therefore
\[ \int_X f_n \, d\mu \;\geq\; \int_{E_n} f_n \, d\mu \;\geq\; c \int_{E_n} s \, d\mu. \]
The set function \(A \mapsto \int_A s \, d\mu\) is a measure, so continuity from below gives \(\int_{E_n} s \, d\mu \to \int_X s \, d\mu\). Letting \(n \to \infty\),
\[ \alpha \geq c \int_X s \, d\mu. \]
Now let \(c \to 1\) to obtain \(\alpha \geq \int_X s \, d\mu\), and take the supremum over all simple \(0 \leq s \leq f\). By the definition of the Lebesgue integral,
\[ \int_X f \, d\mu = \sup\left\{ \int_X s \, d\mu : s \text{ simple},\ 0 \leq s \leq f \right\}, \]
so \(\alpha \geq \int_X f \, d\mu\). Combining the two inequalities yields \(\alpha = \int_X f \, d\mu\), that is,
\[ \lim_{n\to\infty} \int_X f_n \, d\mu = \int_X f \, d\mu. \]
4 Example
A central application is the interchange of summation and integration for non-negative functions.
Let \(g_1, g_2, \ldots : X \to [0,\infty)\) be measurable and define the partial sums \(f_n = \sum_{k=1}^n g_k\). Each \(f_n\) is non-negative and measurable, and the sequence is non-decreasing with \(f_n \uparrow \sum_{k=1}^\infty g_k\) pointwise. Applying the monotone convergence theorem,
\[ \int_X \sum_{k=1}^\infty g_k \, d\mu = \lim_{n\to\infty} \int_X \sum_{k=1}^n g_k \, d\mu = \lim_{n\to\infty} \sum_{k=1}^n \int_X g_k \, d\mu = \sum_{k=1}^\infty \int_X g_k \, d\mu, \]
so for non-negative measurable functions one may always integrate a series term by term.
As a concrete instance, take \(X = (0,1)\) with Lebesgue measure and expand the geometric series \(\frac{1}{1-x} = \sum_{k=0}^\infty x^k\). Since \(-\ln x \geq 0\) on \((0,1)\), every term \(x^k(-\ln x)\) is non-negative, so the theorem justifies
\[ \int_0^1 \frac{-\ln x}{1 - x} \, dx = \sum_{k=0}^\infty \int_0^1 x^k (-\ln x) \, dx = \sum_{k=0}^\infty \frac{1}{(k+1)^2} = \sum_{m=1}^\infty \frac{1}{m^2} = \frac{\pi^2}{6}. \]