The Monotone and Dominated Convergence Theorems

On "The Need for Measure" we named the deepest defect of the Riemann integral. It was not that one exotic function has no area; it was that a sequence of perfectly nice integrable functions can converge to a limit that Riemann cannot touch — and worse, that even when the limit is fine, the two operations

\lim_{n\to\infty} \int f_n \qquad\text{and}\qquad \int \lim_{n\to\infty} f_n

need not agree. Analysis lives on swapping these: differentiating under the integral sign, integrating a Fourier series term by term, computing an expectation as a limit. A theory where the swap fails at the first hurdle is built on sand.

The Lebesgue integral was built precisely so that here the swap works — not always, but under hypotheses so mild that in practice you almost never think about them. This page is those hypotheses: the two great convergence theorems, Monotone Convergence (MCT) and Dominated Convergence (DCT), that turn "\lim\int = \int\lim" from a rare miracle into the everyday reflex of modern analysis. They are the workhorses — cite them a dozen times a day and never blink.

The Monotone Convergence Theorem

The first theorem is almost free — it falls straight out of the way we defined the integral, as a supremum of simple functions climbing from below.

Let f_1 \le f_2 \le f_3 \le \cdots be measurable with 0 \le f_n, increasing pointwise to a limit f = \lim_n f_n = \sup_n f_n. Then

\int f_n \, d\mu \;\uparrow\; \int f \, d\mu, \qquad\text{i.e.}\qquad \lim_{n\to\infty}\int f_n \, d\mu = \int \Bigl(\lim_{n\to\infty} f_n\Bigr) d\mu.

The only hypotheses are non-negativity and monotone increase — no dominating function, no boundedness.
The identity holds in [0,\infty]: if the right side is +\infty then so is the left, and +\infty = +\infty is a perfectly good conclusion.

Why it is nearly a definition. One direction is instant: monotonicity of the integral gives \int f_n \le \int f_{n+1} \le \int f, so the numbers \int f_n increase and stay \le \int f; their limit L exists and L \le \int f. The reverse, L \ge \int f, is the whole content: since \int f is the supremum of \int\varphi over simple 0 \le \varphi \le f, it suffices to beat every such \varphi. Fix one, and fix c \in (0,1); the sets E_n = \{x : f_n(x) \ge c\,\varphi(x)\} increase up to the whole space (because f_n \uparrow f \ge \varphi > c\varphi), so by continuity of the measure from below

\int f_n \,\ge\, \int_{E_n} f_n \,\ge\, c\int_{E_n}\varphi \;\xrightarrow[n\to\infty]{}\; c\int \varphi.

Let c \uparrow 1, then take the sup over \varphi: L \ge \int f. The one-sided, from-below construction of the integral is exactly what makes the limit and the supremum commute — there is no upper-sum machinery fighting the limit, which is precisely where Riemann's version died.

The single most-used corollary of MCT: for any sequence of non-negative measurable functions g_1, g_2, \dots you may swap the sum and the integral, with no justification beyond non-negativity:

\int \sum_{k=1}^{\infty} g_k \, d\mu \;=\; \sum_{k=1}^{\infty} \int g_k \, d\mu.

Why? The partial sums f_n = \sum_{k=1}^{n} g_k are non-negative and increase to the full series, so MCT applies directly and linearity handles each finite sum: \int f_n = \sum_{k=1}^n \int g_k \uparrow \sum_{k=1}^\infty \int g_k. This is "Tonelli for series" — the reason you can integrate a power series or a Fourier series term by term the moment the terms keep one sign, even if it diverges (both sides are then +\infty together).

The Dominated Convergence Theorem

MCT demands monotonicity — the functions must march upward. But most sequences in the wild are not monotone; they wobble. The Dominated Convergence Theorem drops the monotonicity and asks instead for a single ceiling: one integrable function sitting above all the |f_n| at once.

Let f_n be measurable with f_n \to f pointwise almost everywhere. Suppose there is a single integrable g \ge 0 — the dominating function — with

|f_n(x)| \le g(x) \quad\text{for all } n, \text{ for a.e. } x, \qquad \int g\, d\mu < \infty.

Then f is integrable, and

\lim_{n\to\infty}\int f_n \, d\mu = \int f\, d\mu, \qquad\text{and in fact}\qquad \int |f_n - f|\, d\mu \to 0.

The dominator g must be one function, independent of n, and it must be integrable. That single hypothesis is everything.
No monotonicity and no sign restriction — the f_n may oscillate freely inside the envelope \pm g.

Proof sketch, via Fatou. Fatou's lemma (itself a one-line consequence of MCT applied to \inf_{k\ge n} f_k) says \int \liminf f_n \le \liminf \int f_n for non-negative functions. Apply it to the two non-negative sequences that domination hands you, g + f_n \ge 0 and g - f_n \ge 0:

\int g + \int f \le \liminf\Bigl(\int g + \int f_n\Bigr), \qquad \int g - \int f \le \liminf\Bigl(\int g - \int f_n\Bigr).

Because \int g is finite we may cancel it. The first line gives \int f \le \liminf \int f_n; the second, after flipping the sign of the \liminf, gives \limsup \int f_n \le \int f. Squeezing, \lim \int f_n = \int f. The finiteness of the envelope g is exactly what lets us subtract without meeting \infty - \infty — remove it and the whole argument collapses.

Why the hypotheses cannot be dropped: two escaping sequences

Both theorems buy the swap \lim\int = \int\lim with a hypothesis — monotone or dominated. To feel why the hypothesis is not decoration, meet the two standard sequences that converge to 0 at every point yet keep an integral of 1. In each, mass "escapes" and the naive swap gives the flatly wrong 1 = 0.

The tall spike. On [0,\infty) take f_n = n\cdot\mathbf{1}_{(0,\,1/n)} — a rectangle of height n and width 1/n. At any fixed x > 0, once n > 1/x the spike has slipped to the left of x, so f_n(x) = 0; hence f_n \to 0 pointwise. But its area never moves:

\int_0^\infty f_n = n\cdot\frac1n = 1 \;\text{ for every } n, \qquad\text{while}\qquad \int_0^\infty \lim_n f_n = \int_0^\infty 0 = 0.

Drag the slider and watch the paradox happen: the spike shoots up and narrows, the readout for \int f_n stays glued to 1, and the height at a fixed sample point collapses to 0. The mass does not vanish — it is squeezed into an ever-thinner sliver at the origin. There is no integrable g \ge |f_n|: the smallest ceiling is g(x) \ge \sup_n f_n(x) \approx 1/x near 0, which is not integrable. Domination fails, so DCT has nothing to say — correctly.

The sliding bump. A twin defect where the mass runs off to infinity instead of piling up. On [0,\infty) take f_n = \mathbf{1}_{[n,\,n+1]} — a unit box that slides one step right each time. At any fixed x the box eventually passes it by, so again f_n \to 0 pointwise; yet \int f_n = 1 for all n. Once more no integrable dominator exists (a ceiling above every box would be \ge \mathbf{1}_{[1,\infty)}, of infinite integral). Neither sequence is monotone and neither is dominated — so neither theorem applies, and neither swap is valid. That is the point: the hypotheses are load-bearing.

A case where it works: DCT in action

Lest the counterexamples scare you off — when a dominator does exist, DCT makes limits effortless. Consider, on [0,\infty) with ordinary length,

f_n(x) = \frac{n\,\sin(x/n)}{x\,(1 + x^2)}.

Pointwise, n\sin(x/n) \to x for each fixed x (the classic \sin\theta \approx \theta as \theta = x/n \to 0), so

f_n(x) \;\to\; \frac{x}{x\,(1+x^2)} = \frac{1}{1 + x^2} =: f(x).

|f_n(x)| \le \frac{x}{x\,(1+x^2)} = \frac{1}{1+x^2} = g(x), \qquad \int_0^\infty \frac{dx}{1+x^2} = \frac{\pi}{2} < \infty.

One integrable g, independent of n, dominating every f_n. DCT fires immediately, and we may pass the limit through the integral without a second thought:

\lim_{n\to\infty} \int_0^\infty \frac{n\sin(x/n)}{x(1+x^2)}\, dx \;=\; \int_0^\infty \frac{dx}{1 + x^2} \;=\; \frac{\pi}{2}.

No estimate of the messy left-hand integral was ever needed. This is the everyday power of DCT: spot a dominator, and the limit slides straight inside.

The two theorems are not rivals; they are one family, generated in a chain from a single seed. MCT is the seed — it is nearly the definition of the integral, needing no separate proof. From it, Fatou's lemma is a two-line corollary: apply MCT to the increasing sequence h_n = \inf_{k \ge n} f_k \uparrow \liminf f_k to get \int \liminf f_n \le \liminf \int f_n. And from Fatou, DCT follows in the few lines above, by feeding it g \pm f_n \ge 0. So the ladder is

\text{integral defined from below} \;\Longrightarrow\; \text{MCT} \;\Longrightarrow\; \text{Fatou} \;\Longrightarrow\; \text{DCT}.

Each rung earns its keep. MCT swaps limit and integral when the sequence climbs; Fatou gives a one-sided inequality with no hypothesis at all (only non-negativity); DCT restores full equality the moment an integrable envelope is present. Together they make interchanging limit-and-integral, or sum-and-integral, or derivative-and-integral (differentiating under \int is DCT applied to difference quotients), a routine that undergraduate analysis performs on autopilot — the exact operation Riemann could not safely perform even once.

Five traps around the convergence theorems — every one of them a real exam-day error:

Pointwise convergence ALONE never justifies the swap. Both the tall spike n\,\mathbf{1}_{(0,1/n)} and the sliding bump \mathbf{1}_{[n,n+1]} converge to 0 at every point, yet each has \int f_n = 1 \not\to 0. You always need an extra hypothesis — monotone (MCT) or dominated (DCT).
MCT needs BOTH non-negativity and monotone increase. Drop non-negativity and the conclusion can fail; drop monotonicity and you are simply not in MCT's hypotheses (reach for DCT instead). The theorem is not "any convergent sequence".
The dominator must be a SINGLE integrable g, independent of n. Bounding each f_n by its own ceiling g_n (say |f_n| \le n) is worthless — the tall spike satisfies exactly that. One envelope must cover the whole sequence at once, and it must have finite integral.
"Bounded" is not "dominated". Each sliding bump is bounded by 1, a finite constant — but the constant function 1 is not integrable on [0,\infty). On a space of infinite measure a uniform bound is no dominator at all.
Almost-everywhere convergence is enough. DCT only needs f_n \to f a.e., and domination |f_n| \le g a.e. — the integral is blind to null sets, so failure on a measure-zero set costs nothing. You never need convergence at literally every point.