Convergence theorems

"The Lebesgue integral" built $\int f \, d\mu$ for every measurable function by slicing the range into layers, and said that this integral is meant to behave well under limits. "Measurable functions" supplied the other half: a limit of measurable functions is again measurable, so the expression $\int \lim_n f_n \, d\mu$ always has a meaning. What is still missing is the comparison with $\lim_n \int f_n \, d\mu$. This lesson shows with two examples that the two can differ, then proves three theorems, monotone convergence, Fatou's lemma and dominated convergence, that say when they agree. It uses them to integrate series term by term and to differentiate integrals. The probability part of this track reuses these theorems whenever a limit has to pass through an expectation.

Limits and integrals

Let $f_1, f_2, \ldots$ be measurable functions on a measure space $(\Omega, \mathcal{F}, \mu)$ that converge at every point, $f_n(x) \to f(x)$. This is called pointwise convergence. Two numbers can now be computed. One integrates first and takes the limit afterwards, $\lim_n \int f_n \, d\mu$. The other takes the limit first and integrates afterwards, $\int \lim_n f_n \, d\mu = \int f \, d\mu$. Swapping the limit and the integral means claiming that these two numbers are equal. Many routine calculations use this swap without saying so, for example integrating an infinite series term by term, so it is worth knowing exactly when it is allowed.

It is not always allowed. On $[0, 1]$ with length, take the spike $f_n = n \cdot \mathbf{1}_{(0, 1/n]}$, a column of height $n$ standing on the interval $(0, 1/n]$. At a fixed point such as $x = 0.05$ the values are $f_{10}(0.05) = 10$ and $f_{20}(0.05) = 20$, and then $0$ for every $n \ge 21$, because the column has moved to the left of $0.05$. The same happens at every $x > 0$, and $f_n(0) = 0$ for all $n$, so the pointwise limit is the zero function. Every spike nevertheless encloses the same area:

The integral 1 is not fading away; it is squeezed into a smaller and smaller set and is gone in the limit. A second failure looks different. On the half-line $[0, \infty)$ take the escaping bump $f_n = \mathbf{1}_{[n, n+1]}$, a column of height 1 and width 1 that slides to the right. Every $f_n$ has integral 1, but a fixed point $x$ is left behind once $n > x$, so again the pointwise limit is the zero function and the integral of the limit is 0. Here the mass does not get squeezed; it travels off to infinity. The convergence lab has a page on escaping mass that shows a doubling bet losing its mass in the same way.

The figure draws $f_n$ and its limit for four sequences: these two failures and two sequences for which the swap works, which the next sections explain. It opens on the spike with $n = 5$, where $\int f_n = 1$ and the integral of the limit is 0. Move $n$ and watch that the first number never changes, then switch to the capped function and to the power and watch the integrals of $f_n$ approach the integral of the limit.

Pointwise and uniform convergence

Pointwise convergence only promises that each point settles down eventually; how long that takes may depend on the point. For the spike, the point $x = 0.05$ needs $n \ge 21$ and the point $x = 0.001$ needs $n \ge 1{,}001$. There is no single $n$ after which all points have settled, and the biggest value of the spike, $\sup_x |f_n(x) - 0| = n$, grows without bound.

Uniform convergence asks for one rate for all points: $\sup_x |f_n(x) - f(x)| \to 0$. On a space of finite total mass this is enough to swap, because

In words: the integral of a function cannot exceed its largest value times the total mass of the space. As an example take $f_n(x) = x^2 + x/n$ on $[0, 1]$, which converges uniformly to $x^2$ with $\sup_x |f_n - f| = 1/n$. The integrals are $\int_0^1 f_n \, d\lambda = \tfrac13 + \tfrac{1}{2n}$, which differ from $\tfrac13$ by $\tfrac{1}{2n}$, comfortably within the bound $1/n$; at $n = 10$ the difference is $0.05$ and the bound is $0.1$. The factor $\mu(\Omega)$ matters. On $[0, \infty)$ the functions $\tfrac1n \mathbf{1}_{[0, n]}$ converge uniformly to 0, since their largest value is $1/n$, yet each has integral 1.

Uniform convergence is also far too strong to be the whole story. The powers $x^n$ on $[0, 1]$ tend to 0 at every $x < 1$, but the convergence is not uniform: the curve $x^n$ reaches height $\tfrac12$ at $x = 0.5^{1/n}$, which is $0.7071$ for $n = 2$, $0.9330$ for $n = 10$ and $0.9862$ for $n = 50$, so the largest value on $[0, 1)$ stays 1 for every $n$. The integrals $\tfrac{1}{n+1}$ nevertheless tend to $0$, the integral of the limit. The three theorems below explain such cases, and they ask for much less than uniform convergence.

Monotone convergence

The first theorem needs nothing but monotonicity. A sequence is increasing if $f_n(x) \le f_{n+1}(x)$ for every $x$ and $n$. Monotone convergence theorem: if $0 \le f_1 \le f_2 \le \cdots$ are measurable and $f_n \to f$ pointwise, then

where the right side may be $\infty$. The idea is that an increasing sequence has nothing to lose: every $f_n$ lies below $f$, and the functions only ever add layers. The statement allows infinite values. For example $\mathbf{1}_{[0, n]}$ increases to $\mathbf{1}_{[0, \infty)}$ on the half-line, the integrals $n$ increase to $\infty$, and $\infty$ is the integral of the limit.

The proof is worth reading because it shows what the definition of the integral as a supremum buys.

1. The inequality $\le$. Since $f_n \le f$, monotonicity gives $\int f_n \le \int f$ for every $n$. The integrals $\int f_n$ increase, so they have a limit $L$ (possibly $\infty$), and $L \le \int f$.
2. The inequality $\ge$. By definition $\int f$ is the supremum of $\int \varphi$ over simple functions $0 \le \varphi \le f$, so it is enough to show $L \ge \int \varphi$ for each such $\varphi = \sum_j a_j \mathbf{1}_{A_j}$. Fix one, and fix a number $c$ with $0 < c < 1$; the margin $c$ makes room for a strict inequality below.
3. Let $E_n = \{f_n \ge c\varphi\}$. These sets are measurable and increase with $n$, because $f_n$ increases. They fill up the whole space: where $\varphi(x) = 0$ the point lies in $E_1$, and where $\varphi(x) > 0$ we have $c\varphi(x) < \varphi(x) \le f(x)$, so $f_n(x)$ exceeds $c\varphi(x)$ once $n$ is large.
4. On $E_n$ the function $f_n$ is at least $c\varphi$, so $\int f_n \ge \int c\varphi \mathbf{1}_{E_n} = c \sum_j a_j \, \mu(A_j \cap E_n)$.
5. As $n$ grows, $A_j \cap E_n$ increases to $A_j$, so continuity from below, from "Measuring the size of sets", gives $\mu(A_j \cap E_n) \to \mu(A_j)$. Hence $L \ge c \sum_j a_j \mu(A_j) = c \int \varphi$.
6. Let $c \to 1$ and then take the supremum over $\varphi$: $L \ge \int f$.

An example where the limit function is unbounded: the capped functions $f_n = \min(x^{-1/2}, n)$ on $[0, 1]$ follow $x^{-1/2}$ until the curve reaches height $n$, at $x = 1/n^2$, and are flat at height $n$ to the left of that point. They increase with $n$ to $x^{-1/2}$, which is infinite at $0$ but integrable, and each has integral $n \cdot \tfrac{1}{n^2} + \big(2 - \tfrac{2}{n}\big) = 2 - \tfrac1n$. Monotone convergence gives $\int_0^1 x^{-1/2} \, dx = \lim_n (2 - \tfrac1n) = 2$. This is how an unbounded function is integrated: by bounded functions that grow towards it.

Monotone convergence also explains why "The Lebesgue integral" is consistent: if increasing simple functions $\varphi_n$ converge pointwise to $f$, then $\int \varphi_n \, d\mu \to \int f \, d\mu$ for every such sequence, whichever staircases are chosen.

Series of non-negative functions

A series is the limit of its partial sums, and the partial sums of non-negative functions increase. Applying monotone convergence to them gives the rule for integrating a series term by term: if $g_k \ge 0$ for all $k$, then

with both sides possibly infinite. The partial sums $s_n = g_1 + \cdots + g_n$ have $\int s_n = \sum_{k \le n} \int g_k$ by linearity, and $s_n \uparrow \sum_k g_k$, so letting $n \to \infty$ on both sides is exactly monotone convergence. For the geometric series $\sum_{k \ge 0} x^k = 1/(1 - x)$ on $[0, 1)$ this says $\int_0^1 \frac{dx}{1-x} = \sum_{k \ge 0} \frac{1}{k+1} = \infty$: the area under $1/(1-x)$ is infinite exactly because the harmonic series diverges.

A finite answer needs a different series. On $[0, 1)$,

and every term is non-negative. Since $\int_0^1 \frac{x^{k-1}}{k} \, dx = \frac{1}{k^2}$, the rule gives $\int_0^1 -\frac{\ln(1-x)}{x} \, dx = \sum_k \frac{1}{k^2}$. The sum is the classical value $\pi^2/6 \approx 1.6449$, and the partial sums creep up towards it slowly: $1$, $1.25$, $1.361$, $1.424$, and $1.550$ after ten terms. Integrating the left-hand side numerically gives $1.6449$ as well, so the theorem has turned an integral that is awkward by hand into a familiar sum.

The same rule is already familiar in the special case of counting measure. By "The Lebesgue integral", the integral of $a: \mathbb{N} \to [0, \infty)$ against counting measure is the sum $\sum_k a_k$, so for a table of non-negative numbers $a_{n,k}$ the rule says that adding the rows first and then the columns gives the same total as adding the columns first. Non-negativity cannot be dropped. Take the table with $a_{n,n} = 1$, $a_{n,n+1} = -1$ and all other entries $0$. Every row sums to $1 - 1 = 0$, so adding rows first gives $0$. The first column sums to $1$ and every later column to $-1 + 1 = 0$, so adding columns first gives $1$. The two orders disagree because the entries of the table have both signs and the sum of their absolute values is infinite. The section on dominated convergence explains what replaces non-negativity when signs are mixed.

Fatou's lemma

Without monotonicity, non-negative functions still obey an inequality. To state it, recall the lower limit of a sequence of numbers: $\liminf_n a_n = \lim_{n} \inf_{k \ge n} a_k$, the level that the sequence stays above from some point on, up to arbitrarily small margins. For a sequence of functions, apply it at each point. Fatou's lemma: if $f_n \ge 0$ are measurable, then

In words: integrating the limit never gives more than the limit of the integrals. The proof applies monotone convergence to the lower envelopes of the sequence.

1. Let $g_n = \inf_{k \ge n} f_k$. This is measurable, because measurable functions are closed under infima ("Measurable functions"), and $0 \le g_1 \le g_2 \le \cdots$ increases to $\liminf_n f_n$ by definition.
2. Monotone convergence gives $\int g_n \uparrow \int \liminf_n f_n$.
3. For every $k \ge n$ we have $g_n \le f_k$, so $\int g_n \le \int f_k$, hence $\int g_n \le \inf_{k \ge n} \int f_k$.
4. Let $n \to \infty$. The left side tends to $\int \liminf_n f_n$ and the right side to $\liminf_n \int f_n$.

For the spike the lemma reads $0 \le 1$, and the gap is the squeezed mass. The inequality can also be strict when nothing is squeezed or sent away. Let $f_n = \mathbf{1}_{[0, 1/2]}$ for odd $n$ and $f_n = \mathbf{1}_{[1/2, 1]}$ for even $n$. At every point other than $\tfrac12$ the values alternate between 1 and 0, so the lower limit is 0 and the left side is 0, while each $\int f_n = \tfrac12$. The mass is not lost along the way; it just fails to stay at any single point, and the lower limit only sees what survives at every step. For the capped functions and the powers the two sides are equal: $2 = 2$ and $0 = 0$.

The lesson of Fatou's lemma is a direction: in a limit, mass can disappear, but it cannot appear. That is the only statement that holds for every sequence of non-negative functions with no further assumption, and it is the engine behind the next theorem.

Dominated convergence

Both theorems so far need non-negative functions. The next one allows any sign and asks for a single integrable bound. Dominated convergence theorem: if $f_n \to f$ almost everywhere, and there is one function $g$ with $|f_n| \le g$ for all $n$ and $\int g \, d\mu < \infty$, then $f$ is integrable and

Here "almost everywhere" has the meaning of "Lebesgue measure and null sets": the points where convergence fails form a null set, which the integral cannot see. The idea is that a single integrable $g$ keeps the mass from squeezing or escaping, since there is not enough room under $g$ for it. The proof applies Fatou's lemma twice, using the lower limit above and its mirror image $\limsup_n a_n = \lim_n \sup_{k \ge n} a_k$.

1. Since $|f| \le g$ almost everywhere, $f$ is integrable. Both $g + f_n$ and $g - f_n$ are non-negative.
2. Fatou's lemma for $g + f_n \to g + f$ gives $\int (g + f) \le \liminf \int (g + f_n) = \int g + \liminf \int f_n$. Subtracting the finite number $\int g$ leaves $\int f \le \liminf \int f_n$.
3. Fatou's lemma for $g - f_n \to g - f$ gives $\int g - \int f \le \int g - \limsup \int f_n$, so $\limsup \int f_n \le \int f$.
4. Now $\int f \le \liminf \int f_n \le \limsup \int f_n \le \int f$, so the limit exists and equals $\int f$.

The powers $x^n$ on $[0, 1]$ converge to 0 except at the single point $x = 1$, a null set, and all of them lie below the integrable constant 1. The theorem gives $\int_0^1 x^n \, dx = \tfrac{1}{n+1} \to 0$, which is what the explicit formula says, but now it comes without a formula. The theorem applies when there is no formula at all: the functions $(1 + x/n)^{-n}$ lie between 0 and 1 on $[0, 1]$ and converge to $e^{-x}$, so their integrals tend to $1 - e^{-1} \approx 0.6321$; the explicit values are $0.6399$ for $n = 10$, $0.6329$ for $n = 100$ and $0.6322$ for $n = 1{,}000$. With a constant bound on a space of finite total mass the theorem is often called bounded convergence.

The spike has no integrable bound, and the reason shows why one function for all $n$ is required. Any $g$ above every $f_n$ must be at least $\sup_n f_n$. At a point $x$ the spikes with $n \le 1/x$ have value $n$, so this supremum is $\lfloor 1/x \rfloor$, the integer part of $1/x$, which behaves like $1/x$ near 0. Its integral is a harmonic series. The largest of the first $N$ spikes is at least $n$ exactly on $(0, 1/n]$, a set of length $1/n$, so by the layer-cake formula of "The Lebesgue integral" its integral is $\sum_{n=1}^{N} \tfrac1n$, which is $2.929$ for $N = 10$, $5.187$ for $N = 100$ and $7.485$ for $N = 1{,}000$, growing without end. Each spike is bounded on its own, by $n$; what fails is a bound that serves all $n$ at once.

Integrals that depend on a parameter

Many integrals depend on a parameter, $F(t) = \int f(x, t) \, d\mu(x)$, and one often needs $F$ to be continuous or differentiable in $t$. Both questions are limit questions. $F$ is continuous at $t$ when $F(t_n) \to F(t)$ along every sequence $t_n \to t$, which is a swap of a limit and an integral, so dominated convergence applies.

Continuity. If $t \mapsto f(x, t)$ is continuous for almost every $x$, and $|f(x, t)| \le g(x)$ for all $t$ in an interval, with $\int g \, d\mu < \infty$, then $F$ is continuous on that interval. To prove it, take $t_n \to t$: then $f(x, t_n) \to f(x, t)$ for almost every $x$, all dominated by $g$, and dominated convergence gives $F(t_n) \to F(t)$.

Differentiation. If $\partial f/\partial t$ exists and $|\partial f / \partial t (x, t)| \le g(x)$ for all $t$ in an interval, with $g$ integrable, then $F$ is differentiable there and

The proof looks at difference quotients. For a sequence $h_n \to 0$, the quotient $\big(f(x, t + h_n) - f(x, t)\big)/h_n$ tends to $\partial f/\partial t (x, t)$, and by the mean value theorem it equals $\partial f/\partial t (x, \xi)$ for some $\xi$ between $t$ and $t + h_n$, so it is bounded by $g(x)$. Dominated convergence turns the pointwise limit into a limit of integrals, and the integrals of the quotients are the quotients $\big(F(t + h_n) - F(t)\big)/h_n$.

As an example take $F(t) = \int_0^1 e^{tx} \, dx = \frac{e^t - 1}{t}$, with $F(0) = 1$. For $t \in [-1, 1]$ the derivative in $t$ is $x e^{tx}$, which is at most $e$ on $[0, 1]$, an integrable constant. So $F'(0) = \int_0^1 x \, dx = \tfrac12$. A direct check agrees: $F(0.001) = 1.0005002$ and $F(-0.001) = 0.9995002$, so the difference quotient over a step of $0.002$ is $0.5000000$ to seven digits, and the series $F(t) = 1 + \tfrac{t}{2} + \tfrac{t^2}{6} + \cdots$ has slope $\tfrac12$ at 0.

The spike is the standard warning. Let $f(x, t) = \tfrac1t \mathbf{1}_{(0, t]}(x)$ for $t \in (0, 1]$ and $f(x, 0) = 0$. Then $F(t) = 1$ for every $t > 0$, but $F(0) = 0$, so $F$ jumps at 0 although $f(x, t) \to f(x, 0)$ for every $x$. There is no integrable $g$ that bounds all these functions, and continuity fails.

Choosing the right theorem

Each theorem trades an assumption for a conclusion. When a limit has to be moved through an integral, ask the questions in this order.

• The functions are non-negative and increasing: monotone convergence gives equality, even when the value is infinite. This is the tool for series of non-negative terms and for integrals of unbounded functions.
• The functions are non-negative, with nothing known about monotonicity or bounds: Fatou's lemma gives one inequality, $\int \liminf f_n \le \liminf \int f_n$, which is often all that is needed.
• The functions have mixed signs, or are not monotone: look for a single integrable function $g$ above all $|f_n|$. If one exists, dominated convergence gives equality; on a space of finite total mass a constant bound is enough.
• None of these applies: the swap may fail. Look for mass that is squeezed into a shrinking set, as in the spike, or that runs off to infinity, as in the escaping bump. A later lesson of this track names the condition that rules both out exactly.

Summary

• Pointwise convergence $f_n \to f$ does not give $\int f_n \to \int f$: mass can be squeezed into a shrinking set, as in the spike, or run off to infinity, as in the escaping bump.
• Uniform convergence is enough on a space of finite total mass, but it is far too strong: $x^n$ on $[0, 1]$ is not uniformly convergent, yet its integrals converge to the integral of the limit.
• Monotone convergence: for increasing non-negative functions $\int f_n \uparrow \int f$, also when infinite; it gives term-by-term integration of non-negative series, for instance $\sum 1/k^2 = \pi^2/6$.
• Fatou's lemma: $\int \liminf f_n \le \liminf \int f_n$ for non-negative functions; in a limit mass can be lost but never created.
• Dominated convergence: if $f_n \to f$ almost everywhere and $|f_n| \le g$ for one integrable $g$, then $\int f_n \to \int f$; the bound must serve all $n$ at once, and the spike has none.
• The same theorem makes $F(t) = \int f(x, t) \, d\mu$ continuous and lets the derivative pass inside the integral when $f$ or $\partial f/\partial t$ has an integrable bound.