The monotone convergence theorem

Theorem

The monotone convergence theorem

T-085Home MU-302Threads space · chance

Statement

Let \( (X,\mathcal{M},\mu) \) be a measure space and let \( (f_n)_{n=1}^{\infty} \) be a sequence of \( \mathcal{M} \)-measurable functions \( f_n : X \to [0,\infty] \) such that \( 0 \le f_1(x) \le f_2(x) \le \cdots \) for \( \mu \)-almost every \( x \in X \), and \( f_n(x) \to f(x) \) pointwise (equivalently \( f(x) = \sup_n f_n(x) \)) for \( \mu \)-almost every \( x \). Then \( f \) is measurable and \[ \lim_{n\to\infty} \int_X f_n \, d\mu \;=\; \int_X f \, d\mu \;=\; \int_X \lim_{n\to\infty} f_n \, d\mu, \] where both sides are permitted to equal \( +\infty \).

Why it matters

The monotone convergence theorem (MCT) is the load-bearing convergence theorem of Lebesgue integration: it is the first result that lets you exchange a limit and an integral without any domination hypothesis, and every other major limit theorem in the theory — Fatou's lemma, the dominated convergence theorem, the construction of the integral of a general measurable function, countable additivity of the integral over a measure, and Tonelli's theorem for product measures — is proved either directly from MCT or from a chain of results that begins with it.

Its real significance is architectural. The Lebesgue integral of a non-negative measurable function is defined as a supremum over simple functions below it; MCT is what upgrades that static definition into a working calculus by showing the supremum is compatible with pointwise limits. Without it, "integral" would be a number attached to each function individually, with no guarantee that approximating a function well (via simple functions, via truncation, via series) approximates its integral well.

Hypotheses

Non-negativity: \( f_n \ge 0 \) for all \( n \). Drop it: take \( f_n = -\mathbf{1}_{[n,\infty)} \) on \( (\mathbb{R}, \mathcal{L}, \text{Lebesgue measure}) \). Then \( f_n \uparrow 0 \) pointwise, but \( \int f_n = -\infty \) for every \( n \) while \( \int 0 = 0 \); the limit of the integrals is \( -\infty \ne 0 \).

Measurability of each \( f_n \) (with respect to the same \( \sigma \)-algebra \( \mathcal{M} \)). Without it \( \int f_n \, d\mu \) is not even defined, so the statement has no content; the theorem's proof also uses measurability of \( f \) as the pointwise limit of measurable functions, which itself requires each \( f_n \) measurable.

Monotonicity: \( f_n(x) \le f_{n+1}(x) \) for (almost) every \( x \), for every \( n \). Drop it, keeping non-negativity: on \( ([0,1], \mathcal{L}, \text{Lebesgue}) \) let \( f_n = n\,\mathbf{1}_{[0,1/n]} \). Then \( f_n \to 0 \) pointwise everywhere, but \( \int f_n = 1 \) for every \( n \), so \( \lim \int f_n = 1 \ne 0 = \int \lim f_n \). Mass "escapes to a point" instead of being trapped by an increasing envelope.

The hypotheses need only hold \( \mu \)-almost everywhere, not everywhere. This is not optional slack but a genuine strengthening that requires proof: if monotonicity fails on a \( \mu \)-null set \( N \), redefine \( \tilde f_n = f_n \mathbf{1}_{X \setminus N} \), which now increases everywhere to \( \tilde f = f \mathbf{1}_{X\setminus N} \), and \( \int f_n\,d\mu = \int \tilde f_n \, d\mu \), \( \int f\,d\mu = \int \tilde f\,d\mu \) because integrals ignore null sets (a fact itself proved from the definition of \( \int \) as a supremum over simple functions, since altering a function on a null set does not change which simple functions lie beneath it up to a null set). Completeness of \( \mu \) is not required for this step; only that \( N \in \mathcal{M} \) with \( \mu(N)=0 \).

Proof

\text{Assume WLOG that } f_n(x) \le f_{n+1}(x) \text{ for every } x \in X, \text{ every } n.

By the a.e. reduction above: replace \( f_n \) by \( f_n \mathbf{1}_{X \setminus N} \) on the countable union \( N \) of the exceptional null sets (a countable union of \( \mu \)-null sets is \( \mu \)-null, by countable subadditivity). Integrals are unaffected. A

f(x) := \lim_{n\to\infty} f_n(x) = \sup_n f_n(x) \text{ exists in } [0,\infty] \text{ for every } x, \text{ and } f \text{ is } \mathcal{M}\text{-measurable}.

A monotone sequence in \( [0,\infty] \) converges (possibly to \( +\infty \)) to its supremum. Measurability: \( \{x : f(x) \gt a\} = \bigcup_n \{x : f_n(x) \gt a\} \in \mathcal{M} \) for every \( a \in \mathbb{R} \), a countable union of measurable sets, so \( f \) is measurable by the definition of measurability via preimages of \( (a,\infty] \). A

\int_X f_n \, d\mu \text{ is a non-decreasing sequence in } [0,\infty], \text{ so } L := \lim_{n\to\infty}\int_X f_n\,d\mu \text{ exists in } [0,\infty].

Monotonicity of the integral for non-negative measurable functions: \( f_n \le f_{n+1} \) pointwise implies every simple function \( 0 \le \varphi \le f_n \) also satisfies \( \varphi \le f_{n+1} \), so the defining supremum for \( \int f_n \) is taken over a subset of the one for \( \int f_{n+1} \), giving \( \int f_n \,d\mu \le \int f_{n+1}\,d\mu \). A non-decreasing sequence in \( [0,\infty] \) converges to its supremum. A

L \le \int_X f \, d\mu.

Since \( f_n \le f \) for every \( n \) (as \( f = \sup_k f_k \ge f_n \)), monotonicity of the integral (same lemma as Step 3) gives \( \int f_n\,d\mu \le \int f\,d\mu \) for every \( n \); pass to the limit \( n \to \infty \) on the left. A

\text{Fix a simple measurable function } \varphi \text{ with } 0 \le \varphi \le f, \text{ and fix } c \in (0,1). \text{ Define } E_n := \{x \in X : f_n(x) \ge c\,\varphi(x)\}.

Preparation for the reverse inequality \( L \ge \int f\,d\mu \), which cannot be obtained from monotonicity alone since \( f_n \le f \) only bounds one direction. Each \( E_n \in \mathcal{M} \) because \( f_n - c\varphi \) is measurable (difference of measurable functions, finite-valued where \( \varphi \) is a simple function with finite values on a partition) and \( E_n = \{f_n - c\varphi \ge 0\} \). B

E_1 \subseteq E_2 \subseteq E_3 \subseteq \cdots \quad \text{and} \quad \bigcup_{n=1}^\infty E_n = X.

Monotonicity \( f_n \le f_{n+1} \) gives \( E_n \subseteq E_{n+1} \) directly from the definition. For the union: if \( \varphi(x) = 0 \) then \( x \in E_1 \) trivially. If \( \varphi(x) \gt 0 \) then \( c\varphi(x) \lt \varphi(x) \le f(x) = \sup_n f_n(x) \), so by definition of supremum some \( f_n(x) \gt c\varphi(x) \), i.e. \( x \in E_n \). B

\int_X f_n \, d\mu \;\ge\; \int_{E_n} f_n \, d\mu \;\ge\; c \int_{E_n} \varphi \, d\mu.

First inequality: \( f_n \ge 0 \) on \( X \setminus E_n \), so restricting the domain of integration of a non-negative function cannot increase the integral (monotonicity again, applied to \( f_n \mathbf{1}_{E_n} \le f_n \)). Second inequality: on \( E_n \), \( f_n \ge c\varphi \) by definition of \( E_n \), and \( c\varphi \) restricted to \( E_n \) is again a non-negative simple function, so \( \int_{E_n} f_n\,d\mu \ge \int_{E_n} c\varphi\,d\mu = c\int_{E_n}\varphi\,d\mu \). B

\int_{E_n} \varphi \, d\mu \;\longrightarrow\; \int_X \varphi \, d\mu \quad \text{as } n \to \infty.

Write \( \varphi = \sum_{i=1}^k a_i \mathbf{1}_{A_i} \) in canonical form ( \( a_i \ge 0 \), \( A_i \) disjoint measurable). Then \( \int_{E_n}\varphi\,d\mu = \sum_i a_i\, \mu(A_i \cap E_n) \), and \( A_i \cap E_n \uparrow A_i \) as \( n\to\infty \) (Step 6), so by continuity from below of the measure \( \mu \) (a direct consequence of countable additivity applied to the disjoint decomposition \( A_i = (A_i\cap E_1) \cup \bigcup_{n\ge 1}(A_i \cap E_{n+1}\setminus E_n) \)), \( \mu(A_i\cap E_n) \to \mu(A_i) \) for each \( i \); the finite sum converges term by term. C

L \;\ge\; c \int_X \varphi \, d\mu.

Let \( n \to \infty \) in Step 7: the left side tends to \( L \) by Step 3, the right side tends to \( c\int_X\varphi\,d\mu \) by Step 8 (multiplying a convergent sequence by the constant \( c \)). B

L \;\ge\; \int_X \varphi\,d\mu \quad \text{for every simple } 0\le\varphi\le f.

Let \( c \uparrow 1 \) in Step 9; the left side \( L \) does not depend on \( c \), so \( L \ge c\int\varphi\,d\mu \) for all \( c\in(0,1) \) forces \( L \ge \int_X\varphi\,d\mu \). A

L \;\ge\; \int_X f \, d\mu.

By the definition of the Lebesgue integral of a non-negative measurable function, \( \int_X f\,d\mu = \sup\{\int_X\varphi\,d\mu : \varphi \text{ simple, measurable}, 0\le\varphi\le f\} \); Step 10 shows \( L \) is an upper bound for this same set, hence dominates its supremum. A

L = \int_X f\,d\mu.

Combine Step 4 ( \( L \le \int f\,d\mu \) ) with Step 11 ( \( L \ge \int f\,d\mu \) ) and antisymmetry of \( \le \) on \( [0,\infty] \). Unwinding \( L = \lim_n \int f_n\,d\mu \) gives the theorem. A

Result

f_n \uparrow f,\ f_n \ge 0 \text{ measurable} \quad\Longrightarrow\quad \lim_{n\to\infty}\int_X f_n\,d\mu = \int_X f\,d\mu

Reading. If you build up a non-negative function as an increasing tower of measurable approximations, the total area under the tower converges to exactly the area under the final function — no mass can leak away and none can be manufactured, even when both sides are infinite.

Scope. Holds on any measure space \( (X,\mathcal{M},\mu) \), for any \( \sigma \)-additive (possibly infinite, possibly non-\( \sigma \)-finite) measure \( \mu \), with values allowed to be \( +\infty \) on either side. It requires non-negativity and monotonicity of the sequence (a.e. suffices); it does not require continuity, finiteness of \( \mu(X) \), or any domination by an integrable function.

Corollaries & converses

Term-by-term integration of series. If \( g_k \ge 0 \) measurable for \( k = 1,2,\dots \), then \( \int_X \sum_{k=1}^\infty g_k\,d\mu = \sum_{k=1}^\infty \int_X g_k\,d\mu \), by applying MCT to the partial sums \( f_n = \sum_{k=1}^n g_k \), which increase to \( \sum_k g_k \). This is the standard route to countable additivity of \( E \mapsto \int_E f\,d\mu \).
Fatou's lemma follows from MCT (not the reverse route usually taken in textbooks, but a valid one): apply MCT to \( h_n = \inf_{k\ge n} f_k \), which increases to \( \liminf_n f_n \), and use \( \int h_n \le \int f_n \) for all indices past \( n \).
Existence of the integral as a well-defined limit along any increasing sequence of simple functions approximating a measurable \( f \ge 0 \): different approximating sequences give the same limiting integral, which is what makes "\( \int f\,d\mu \)" unambiguous in the first place.
Converse fails. The conclusion \( \lim_n\int f_n\,d\mu = \int f\,d\mu \) can hold for sequences that are neither monotone nor non-negative (e.g. by dominated convergence, or by luck); it does not imply the sequence was increasing. MCT gives a sufficient, not necessary, condition for exchanging limit and integral.
Decreasing analogue is false in general. There is no "monotone decreasing convergence theorem" without an extra integrability hypothesis on the first term — see Fails without, below.

Fails without

Non-negativity dropped: \( f_n = -\mathbf{1}_{[n,\infty)} \) on \( \mathbb{R} \) increases to \( f \equiv 0 \), yet \( \int f_n\,d\mu = -\infty \ne 0 = \int f\,d\mu \) for every \( n \).
Monotonicity dropped: the "typewriter" / escaping-bump sequence \( f_n = n\,\mathbf{1}_{[0,1/n]} \) on \( [0,1] \) converges pointwise to \( 0 \) with \( \int f_n = 1 \) for all \( n \), so \( \lim\int f_n = 1 \ne 0 \).
Decreasing sequences without an integrable dominator: \( f_n = \mathbf{1}_{[n,\infty)} \) on \( (\mathbb{R}, \text{Lebesgue}) \) decreases to \( f \equiv 0 \), but \( \int f_n\,d\mu = \infty \) for every \( n \), so \( \lim\int f_n = \infty \ne 0 = \int f \). (The correct decreasing statement needs \( \int f_1\,d\mu \lt \infty \); this is a separate theorem, not a corollary of MCT applied blindly to \( -f_n \), since \( -f_n \) is not non-negative.)
Measurability dropped: if \( f_n \) is a non-measurable, non-negative, increasing sequence (constructible using a non-measurable set via the axiom of choice), \( \int f_n\,d\mu \) is undefined at every stage, so the equation cannot even be stated.

Common errors

Applying MCT to a decreasing sequence directly, forgetting the extra hypothesis \( \int f_1 \lt \infty \) that the decreasing version requires (see the third counterexample above).
Forgetting the non-negativity hypothesis and applying MCT to signed \( f_n \uparrow f \); students often try to "shift" by adding a constant to force non-negativity, but this is illegal when \( \mu(X) = \infty \) since \( \int_X c\,d\mu = \infty \) swamps the argument.
Confusing "monotone in \( n \) for each fixed \( x \)" with "monotone as functions", e.g. assuming \( \|f_n\| \) increasing suffices; the pointwise (a.e.) order is what is needed, not any global norm comparison.
Treating MCT as justifying \( \lim_n \int f_n = \int \lim_n f_n \) for arbitrary increasing sequences of Riemann integrals; MCT is a Lebesgue-theory statement and the Riemann integral is not even guaranteed to exist for the limit function even when each \( f_n \) is Riemann integrable and increasing.
Misapplying the theorem when the limit is only assumed to exist in \( [0,\infty] \) but the "monotone" hypothesis is checked only on a dense subset of \( X \) rather than \( \mu \)-a.e.; density is not the same as full or almost-full measure unless \( \mu \) has full support in a compatible way.

Discussion

MCT is where Lebesgue's construction earns its keep relative to Riemann's. In Riemann integration, an increasing sequence of Riemann-integrable functions bounded above by another Riemann-integrable function need not have a Riemann-integrable limit (the limit can fail to be Riemann integrable even when it is bounded, e.g. an enumeration of the rationals in \( [0,1] \) building up the Dirichlet function). Lebesgue's theory sidesteps this because integrability of the limit is automatic — measurability is preserved under pointwise limits (Step 2), and MCT then hands you the value of the integral for free.

Historically the result appears in Lebesgue's 1902 thesis in a form tailored to his construction of the integral via level sets, and was later isolated as a standalone convergence theorem once the abstract measure-theoretic framework of Carathéodory and others made the argument portable to any measure space, not just Lebesgue measure on \( \mathbb{R}^n \). Its abstract form (stated here) is what makes probability theory possible as a special case: with \( \mu \) a probability measure, MCT becomes the statement that expectation commutes with increasing limits of non-negative random variables, which underlies the construction of expectations for general random variables (via \( X = X^+ - X^- \)) and the definition of conditional expectation for non-negative variables via monotone limits of simple ones.

The proof structure — squeeze the integral of the limit between an easy upper bound (monotonicity) and a harder lower bound built from simple functions and continuity of measure from below — is a template reused throughout the subject. Fatou's lemma, the dominated convergence theorem's proof via Fatou applied twice, and the construction of product measures via Tonelli's theorem all repeat some version of "approximate by simple functions, use continuity of \( \mu \) from below, take a supremum."

A subtlety worth flagging: the theorem is often stated with \( f_n \uparrow f \) "in the sense of a.e. convergence to \( \sup_n f_n \)", but the more primitive and slightly more general statement only assumes \( (f_n) \) is a pointwise a.e. non-decreasing sequence and defines \( f := \lim_n f_n \) directly, without separately asserting \( f = \sup_n f_n \); these coincide by monotone convergence of extended-real sequences, but conflating "the sequence has a limit" with "the sequence is bounded" is a common source of confusion when \( f \) is allowed to take the value \( +\infty \) on a set of positive measure — the theorem does not require \( \mu(\{f = \infty\}) = 0 \), and indeed \( \int_X f\,d\mu \) may itself equal \( +\infty \), which is a perfectly valid conclusion, not a failure of the theorem.

Worked examples

\text{Compute } \int_0^\infty e^{-x}\,dx \text{ (Lebesgue integral on } (\mathbb{R}, \mathcal{L}, \text{Lebesgue measure)) via an increasing sequence.}

Set up: let \( f_n(x) = e^{-x}\mathbf{1}_{[0,n]}(x) \) for \( n = 1,2,\dots \). A

0 \le f_n \le f_{n+1}, \qquad f_n(x) \uparrow e^{-x}\mathbf{1}_{[0,\infty)}(x) \text{ for every } x \in \mathbb{R}.

Each \( f_n \) is non-negative, measurable (product of a continuous function and an indicator of a Borel set), and \( \mathbf{1}_{[0,n]} \le \mathbf{1}_{[0,n+1]} \) pointwise, so \( f_n \le f_{n+1} \). For fixed \( x \ge 0 \), once \( n \ge x \) we have \( f_n(x) = e^{-x} \), so \( f_n(x) \to e^{-x} \); for \( x \lt 0 \), \( f_n(x) = 0 \) for all \( n \). A

\int_{\mathbb{R}} f_n\,d\mu = \int_0^n e^{-x}\,dx = 1 - e^{-n}

For fixed finite \( n \), \( f_n \) is a bounded function supported on a bounded interval, so its Lebesgue integral coincides with the (elementary) Riemann integral \( \int_0^n e^{-x}\,dx \), computed by the fundamental theorem of calculus. A

\int_{\mathbb{R}} e^{-x}\mathbf{1}_{[0,\infty)}\,d\mu = \lim_{n\to\infty}\int_{\mathbb{R}} f_n\,d\mu = \lim_{n\to\infty}(1-e^{-n}) = 1.

Apply MCT to \( (f_n) \) using Step 2 as the hypothesis; the theorem licenses passing the limit inside the integral, then Step 3 gives an explicit elementary limit. B

\int_0^\infty e^{-x}\,dx = 1

Reading. MCT justifies computing an improper integral over an unbounded domain by exhausting it with bounded pieces and passing to the limit — exactly the manoeuvre used informally in calculus, now made rigorous.

\text{Show } \sum_{k=1}^\infty \frac{1}{2^k}\,\mathbf{1}_{[0,k]}(x) \text{ integrates to } \sum_{k=1}^\infty \frac{k}{2^k} \text{ on } ([0,\infty), \text{Lebesgue measure)}.

Set up: let \( g_k(x) = \frac{1}{2^k}\mathbf{1}_{[0,k]}(x) \ge 0 \), measurable for each \( k \), and let \( f_n = \sum_{k=1}^n g_k \). A

f_n \le f_{n+1} \quad (\text{since } f_{n+1} - f_n = g_{n+1} \ge 0), \qquad f_n(x) \uparrow \sum_{k=1}^\infty g_k(x) =: g(x) \text{ for every } x.

Finite sums of non-negative terms are non-decreasing as terms are added; the pointwise limit of the partial sums is by definition the series value (in \( [0,\infty] \), always well-defined for non-negative terms since the partial sums are monotone). A

\int_0^\infty f_n \, dx = \sum_{k=1}^n \int_0^\infty g_k\,dx = \sum_{k=1}^n \frac{k}{2^k}

Linearity of the integral for finite sums of non-negative measurable functions (itself a consequence of the definition via simple functions, proved before MCT and used here, not by MCT); each \( \int_0^\infty g_k\,dx = \frac{1}{2^k}\cdot k \) since \( g_k \) is \( 2^{-k} \) on an interval of length \( k \). A

\int_0^\infty g\,dx = \lim_{n\to\infty}\int_0^\infty f_n\,dx = \sum_{k=1}^\infty \frac{k}{2^k} = 2.

Apply MCT to \( (f_n) \) via Step 2, then evaluate the classical series \( \sum_{k\ge1} k x^k = x/(1-x)^2 \) at \( x = 1/2 \), giving \( (1/2)/(1/4) = 2 \). B

\int_0^\infty \sum_{k=1}^\infty \frac{\mathbf{1}_{[0,k]}(x)}{2^k}\,dx \;=\; \sum_{k=1}^\infty \int_0^\infty \frac{\mathbf{1}_{[0,k]}(x)}{2^k}\,dx \;=\; 2

Reading. This is the series corollary of MCT in action: integrating a non-negative series term by term and summing agrees with summing the series first and integrating, with no extra hypotheses beyond non-negativity of each term.

Problems

Let \( f_n(x) = x^n \) on \( ([0,1], \mathcal{L}, \text{Lebesgue}) \). Determine whether \( (f_n) \) is monotone, find the pointwise limit, and use MCT (or explain why it does not apply) to compute \( \lim_n \int_0^1 x^n\,dx \).
Solution
For \( x \in [0,1] \), \( x^{n+1} \le x^n \): the sequence is monotone decreasing, not increasing, so MCT (increasing version) does not directly apply. However \( g_n := 1 - f_n = 1-x^n \) is non-negative (since \( x^n\le1 \)) and increasing in \( n \) (as \( x^n \) decreases), with \( g_n \uparrow 1 - \mathbf{1}_{\{1\}}(x)\cdot 1\)... more simply \( g_n(x) \to 1 \) for \( x\in[0,1) \) and \( g_n(1) = 0 \) for all \( n \), so \( g_n \uparrow g \) where \( g = \mathbf{1}_{[0,1)} \). By MCT, \( \int_0^1 g_n\,dx \to \int_0^1 g\,dx = 1 \) (since \( \{1\} \) has measure zero, \( \int_0^1 \mathbf{1}_{[0,1)} = 1 \)). Since \( \int_0^1 g_n\,dx = 1 - \int_0^1 x^n\,dx = 1 - \frac{1}{n+1} \), matching the limit \( 1 \) confirms \( \int_0^1 x^n\,dx = \frac{1}{n+1} \to 0 \), consistent with direct computation. The point of the exercise is that MCT was applied to \( 1-f_n \), not to \( f_n \) itself.
Give an example of a sequence of non-negative measurable functions \( f_n \) on \( (\mathbb{R}, \mathcal{L}) \) that converges pointwise everywhere to \( 0 \), is not monotone, yet still satisfies \( \int f_n\,d\mu \to \int 0\,d\mu = 0 \). Explain why this does not contradict the necessity of monotonicity in MCT.
Solution
Take \( f_n = \frac{1}{n}\mathbf{1}_{[0,1]} \). Then \( f_n \to 0 \) pointwise, \( f_n \) is not monotone in the required sense in general (here it happens to be decreasing, not increasing, so still not covered by MCT), and \( \int f_n\,d\mu = \frac{1}{n} \to 0 \), matching \( \int 0\,d\mu = 0 \). This is consistent with the theorem because MCT states a sufficient condition for \( \lim\int f_n = \int\lim f_n \), not a necessary one (see Corollaries: converse fails); other tools (e.g. dominated convergence with dominator \( \mathbf{1}_{[0,1]} \)) certify the limit here instead.
Let \( \mu \) be counting measure on \( \mathbb{N} \), and let \( f_n : \mathbb{N} \to [0,\infty] \) be defined by \( f_n(k) = \left(1 - \frac{1}{n}\right)^k \) for \( k \le n \) and \( f_n(k) = 0 \) for \( k \gt n \)... instead use the cleaner monotone example \( f_n(k) = \mathbf{1}_{\{1,\dots,n\}}(k)\cdot \frac{1}{k^2} \). Show \( f_n \uparrow f \) for an explicit \( f \), and evaluate \( \lim_n \sum_{k=1}^\infty f_n(k) \) using MCT, identifying the sum with a known series.
Solution
Here \( \int_{\mathbb{N}} f_n \, d\mu = \sum_{k=1}^n \frac{1}{k^2} \) (integration against counting measure is summation). Since \( \mathbf{1}_{\{1,\dots,n\}} \le \mathbf{1}_{\{1,\dots,n+1\}} \) pointwise and \( 1/k^2 \ge 0 \), \( (f_n) \) is non-negative and increasing, with \( f_n(k) \to \frac{1}{k^2} \) for every fixed \( k \) once \( n \ge k \); so \( f_n \uparrow f \) with \( f(k) = 1/k^2 \). By MCT, \( \lim_n \sum_{k=1}^n \frac{1}{k^2} = \sum_{k=1}^\infty \frac{1}{k^2} = \int_{\mathbb{N}} f\,d\mu \). This recovers the classical fact (via the Basel problem) that the partial sums converge to \( \pi^2/6 \); MCT is what certifies that "the integral of the limit" and "the limit of the integrals" are the same number a priori, before any specific series-evaluation technique is applied.
Let \( (X,\mathcal{M},\mu) \) be a measure space and \( E_1 \subseteq E_2 \subseteq \cdots \) an increasing sequence of measurable sets with \( E = \bigcup_n E_n \). For a fixed non-negative measurable \( f \), prove \( \int_{E_n} f\,d\mu \to \int_E f\,d\mu \) using MCT (this is "continuity of the integral from below" and generalises continuity of \( \mu \) from below, which is the case \( f \equiv 1 \)).
Solution
Let \( f_n = f\cdot\mathbf{1}_{E_n} \). Each \( f_n \ge 0 \) is measurable (product of measurable functions). Since \( E_n \subseteq E_{n+1} \), \( \mathbf{1}_{E_n} \le \mathbf{1}_{E_{n+1}} \) pointwise, so \( f_n \le f_{n+1} \): the sequence is non-negative and increasing. For \( x \in E \), \( x \in E_n \) for all large \( n \) (by definition of the union of an increasing sequence), so \( f_n(x) \to f(x) \); for \( x \notin E \), \( f_n(x) = 0 \) for all \( n \), so \( f_n(x) \to 0 = f(x)\mathbf{1}_E(x) \). Hence \( f_n \uparrow f\mathbf{1}_E \) pointwise everywhere. By MCT, \( \int_X f_n\,d\mu \to \int_X f\mathbf{1}_E\,d\mu \), i.e. \( \int_{E_n} f\,d\mu \to \int_E f\,d\mu \), as required.
(Challenge) Construct an example on \( (\mathbb{R}, \mathcal{L}, \text{Lebesgue}) \) of a sequence of non-negative measurable functions \( f_n \uparrow f \) pointwise everywhere with \( \int f_n\,d\mu \lt \infty \) for every finite \( n \), but \( \int f\,d\mu = \infty \). Confirm this is consistent with MCT rather than a counterexample to it.
Solution
Take \( f_n = \sum_{k=1}^n \mathbf{1}_{[k,k+1)} \) (equivalently \( f_n = \mathbf{1}_{[1,n+1)} \)). Each \( f_n \) is a finite sum of indicators of unit intervals, so \( \int f_n\,d\mu = n \lt \infty \). The sequence increases pointwise (adding one more interval each step) to \( f = \mathbf{1}_{[1,\infty)} \), and \( \int f\,d\mu = \mu([1,\infty)) = \infty \). MCT predicts exactly \( \lim_n \int f_n\,d\mu = \lim_n n = \infty = \int f\,d\mu \): both sides are \( \infty \), and the theorem explicitly allows this (Result/Scope states values in \( [0,\infty] \) are permitted on both sides). There is no contradiction — MCT never asserts the common value is finite, only that the two sides agree in \( [0,\infty] \).