The Lebesgue integral

"Measurable functions" showed which functions allow the question "how much of the space is sent into this range of values?" to be answered, and it introduced the simple functions and the dyadic staircases that approach any measurable function from below. "Lebesgue measure and null sets" supplied the measure itself, length on the line and area in the plane. What is still missing is a way to add up a function over a space: to say what the total of $f$ is when $f$ has a value at every point of $[0, 1]$, of the unit square, or of any other measure space. This lesson builds that total, the Lebesgue integral $\int f\,d\mu$ for an arbitrary measure $\mu$. It starts from the column-by-column method of school calculus, shows where it breaks, and replaces it by a layer-by-layer method that needs only the measures of the level sets of $f$. The next lesson, "Convergence theorems", asks what happens to integrals when functions are replaced by their limits.

Riemann's columns

The integral of school calculus, the Riemann integral, slices the domain. To find the area under $f$ on $[0, 1]$, cut $[0, 1]$ into $N$ columns of width $1/N$, and on each column record the lowest and the highest value of $f$. Rectangles with the lowest heights fit under the graph, and their total area is the lower sum. Rectangles with the highest heights stick out above it, and their total area is the upper sum. The integral is the number that stays squeezed between the two sums as the columns become narrow.

Take the hump $f(x) = 4x(1 - x)$, which rises from 0 at $x = 0$ to 1 at $x = \tfrac12$ and falls back to 0 at $x = 1$. With $N = 4$ columns the lowest values are $0, 0.75, 0.75, 0$ and the highest are $0.75, 1, 1, 0.75$, so the lower sum is $0.375$ and the upper sum is $0.875$. With $N = 8$ the two sums are $0.5313$ and $0.7813$. The gap between them halves every time the number of columns doubles, and both sums close in on the same number, $2/3$.

This works because a continuous function hardly changes on a narrow column. The method quietly assumes that $f$ is nearly constant on short stretches of the domain, and a function that jumps within every stretch breaks the assumption. Consider the indicator $\mathbf{1}_{\mathbb{Q}}$ on $[0, 1]$, equal to 1 at the rational numbers and 0 elsewhere. Every column, however narrow, contains both rational and irrational numbers, since both are dense, so every upper sum is 1 and every lower sum is 0: the Riemann integral does not exist.

Yet "Lebesgue measure and null sets" showed that the rationals have length 0. A function equal to 1 on a set that occupies no room should add nothing to a total. Columns cannot see this, because they look at where a value sits; layers look at how much of the space carries each value.

Lebesgue's layers

Lebesgue explained his idea with coins. A cashier has a heap of mixed coins and wants their total value. One way is to take the coins one by one in the order they come to hand and keep a running sum: that is the column method, which follows the points of the domain in order. The other way is to sort the coins into piles by value, count each pile, and add up value times count. This never needs to know in what order the coins lay; it needs only how many coins there are of each value.

Replace the heap by the space $\Omega$, the coins by points, and the number of coins by the measure. The layer method groups the points of $\Omega$ by the value of $f$, measures each group, and adds up value times measure, whether the points of a group are close together or scattered over the whole space.

Do this for $f(x) = 4x(1 - x)$ with the staircase $\varphi_2$ of "Measurable functions", which rounds every value down to a multiple of $\tfrac14$. It takes the value $0$ on a set of length $0.1340$, the value $\tfrac14$ on a set of length $0.1589$, the value $\tfrac12$ on a set of length $0.2071$ and the value $\tfrac34$ on a set of length $0.5$, and the four lengths add up to 1. The total of the staircase is

This underestimates the area $2/3$, because every value was rounded down; finer bands improve it. For $\mathbf{1}_{\mathbb{Q}}$ the only layer that counts is the value 1 on a set of length 0, contributing $1 \cdot 0 = 0$, so the total is 0, as it should be. Layers need the measure of the sets $\{a \le f < b\}$, and measurability, the subject of "Measurable functions", is exactly what makes these sets measurable.

Integrals of simple functions

The layer method is exact for functions that take finitely many values. Let $\varphi$ be a simple function. Write it in standard form, with distinct values $a_1, \ldots, a_m$ taken on disjoint measurable sets $A_1, \ldots, A_m$. Its integral with respect to a measure $\mu$ is

with the convention $0 \cdot \infty = 0$, so that a value 0 never contributes, even on a set of infinite measure. In words: each value times the amount of space on which it is taken.

The same number results from any way of writing $\varphi$ as a combination of indicators, even if the sets overlap. Take $\varphi = 2\cdot\mathbf{1}_{[0, 0.5]} + 3\cdot\mathbf{1}_{[0.25, 1]}$ on $[0, 1]$ with length as the measure, the example of "Measurable functions". Reading the formula off the overlapping form gives $2 \cdot 0.5 + 3 \cdot 0.75 = 3.25$. The standard form $2\cdot\mathbf{1}_{[0, 0.25)} + 5\cdot\mathbf{1}_{[0.25, 0.5]} + 3\cdot\mathbf{1}_{(0.5, 1]}$ gives $2 \cdot 0.25 + 5 \cdot 0.25 + 3 \cdot 0.5 = 3.25$. The reason is additivity. Cut the space into the pieces on which none of the indicators changes, the atoms of the σ-algebra that these sets generate, as in "σ-algebras and Borel sets". Each set $A_k$ is a disjoint union of atoms, so $\mu(A_k)$ is the sum of the measures of its atoms, and both ways of writing $\varphi$ add up the same terms.

Two rules follow for simple functions $\varphi$ and $\psi$: the integral is linear, $\int (a\varphi + b\psi)\,d\mu = a\int \varphi\,d\mu + b\int \psi\,d\mu$, because the combination is again a list of indicators; and monotone, since $\varphi \le \psi$ means that on every atom the value of $\varphi$ is at most that of $\psi$, so $\int \varphi\,d\mu \le \int \psi\,d\mu$. For an indicator, $\int \mathbf{1}_A\,d\mu = \mu(A)$.

A dart board shows a simple function with several layers. Take the unit square, with a target centred in it. Points closer than $0.1$ to the centre score 10, points between distance $0.1$ and $0.25$ score 5, between $0.25$ and $0.4$ score 2, between $0.4$ and $0.5$ score 1, and the four corners beyond distance $0.5$ score 0. The score is a function $s$ on the square with five values, each taken on a ring or on the corners. The measure is area, the two-dimensional Lebesgue measure from "Lebesgue measure and null sets", under which a disc of radius $r$ has area $\pi r^2$. The integral $\int s \, d\lambda$ is the score of the board weighted by area: every ring contributes its score times its area.

Integrals of non-negative functions

A general function takes infinitely many values, so the sum of value times measure has no finite list to run over. The remedy is to use simple functions as lower estimates. Let $f \ge 0$ be measurable. Every simple function $\varphi$ with $0 \le \varphi \le f$ is an underestimate of the total of $f$, and the best underestimate is the total:

The value lies in $[0, \infty]$, and it can be $\infty$. If $f$ is itself simple, the definition returns the old formula, since $f$ is one of the allowed $\varphi$ and monotonicity bounds all the others.

A supremum over all simple functions below $f$ looks unworkable, but one sequence suffices. The dyadic staircases of "Measurable functions", $\varphi_n = \min\big(n, \lfloor 2^n f \rfloor / 2^n\big)$, increase to $f$ at every point, and their integrals increase to $\int f\,d\mu$. This last step is the monotone convergence theorem, the first result of the next lesson, so here it is taken as given. For $f(x) = 4x(1 - x)$, the set $\{f \ge y\}$ is an interval of length $\sqrt{1 - y}$, as computed in "Measurable functions", and adding up the layers of $\varphi_n$ gives

• $n = 1$: $0.3536$, with a distance of $0.3131$ to the exact value $2/3 = 0.6667$.
• $n = 2$: $0.5183$, as computed above, distance $0.1484$.
• $n = 3$: $0.5956$, distance $0.0710$.
• $n = 4$: $0.6323$, distance $0.0343$; at $n = 6$ the value is $0.6585$.

The distance roughly halves with every refinement, and the integrals of the staircases creep up on $2/3$ from below.

The figure draws the staircase $\varphi_n$ under the curve and, for comparison, the Riemann construction with the same number of pieces. Choose "Slice the range" or "Slice the domain" and move $n$. With the defaults, slicing the range at $n = 3$ uses $8$ levels and shows the Lebesgue sum $0.5956$ next to the exact value $0.6667$. Switching to slicing the domain shows $8$ midpoint columns and the midpoint Riemann sum $0.6719$.

For this smooth hump the midpoint Riemann sums converge faster than the staircase sums: $0.6719$ against $0.5956$ at $n = 3$. Speed on smooth functions is not the point of the Lebesgue construction. The point is that it needs only the measure of level sets, so it works for every measurable function and for every measure.

The layer-cake formula

The staircase can be taken apart into thin flat layers. At a point where $\varphi_n$ has the value $j/2^n$, exactly the indicators $\mathbf{1}\{f \ge k/2^n\}$ for $k = 1, \ldots, j$ are on, and each adds $2^{-n}$. So

a stack of slabs of height $2^{-n}$, the $k$-th slab lying over the set where $f$ reaches the height $k/2^n$. Integrating a sum of simple functions term by term gives

The right side is a sum of rectangles of width $2^{-n}$ whose heights are the values of the tail function $t \mapsto \mu(f > t)$, the measure of the part of the space where $f$ exceeds $t$. This function decreases as $t$ grows, and the sum is an approximation to the area under its graph. Letting $n$ grow, and using that $\int \varphi_n\,d\mu \to \int f\,d\mu$, gives the layer-cake formula:

The integral of $f$ over the space equals the ordinary area under the tail function. The picture is a cake: the layer at height $t$ has the size of the part of the space that reaches it, and the cake is the sum of its layers. The derivation is an outline; the passage to the limit uses monotone convergence.

For the hump, $\mu(f > t) = \sqrt{1 - t}$ for $0 \le t < 1$ and $0$ beyond, so

the same $2/3$ as the column integral $\int_0^1 4x(1 - x)\,dx = 2 - \tfrac43$. For the function $f(x) = 1/x$ on $(0, 1]$, which is unbounded, the set $\{f > t\}$ is all of $(0, 1]$ for $t < 1$ and the interval $(0, 1/t)$ for $t \ge 1$, so $\mu(f > t) = \min(1, 1/t)$, and the area is $1 + \int_1^\infty dt/t = \infty$: the integral exists and equals $\infty$, because the layers shrink too slowly to have a finite area.

On the dart board the layers are easy to see. The score reaches at least 1 on the disc of radius $0.5$, at least 2 on the disc of radius $0.4$, at least 5 on the disc of radius $0.25$ and at least 10 on the disc of radius $0.1$. Their areas are the tail values $0.7854$, $0.5027$, $0.1963$ and $0.0314$. The score climbs from 0 to 1, 2, 5 and 10, so the layers have heights $1, 1, 3, 5$, and weighting each tail area by its layer height must reproduce the integral found ring by ring, as the next worked example confirms.

Functions with both signs

The construction above needs $f \ge 0$. A function with both signs is split into two non-negative parts: the positive part $f^+ = \max(f, 0)$ and the negative part $f^- = \max(-f, 0)$. Then $f = f^+ - f^-$ and $|f| = f^+ + f^-$. The positive part keeps the places where $f$ lies above the axis, and the negative part is the reflection of the places where it lies below.

A function $f$ is integrable if $\int |f|\,d\mu < \infty$, that is, if both $\int f^+\,d\mu$ and $\int f^-\,d\mu$ are finite, and then

The set of integrable functions is written $L^1(\mu)$. As an example, $f(x) = \sin(2\pi x)$ on $[0, 1]$ is positive on $(0, \tfrac12)$ and negative on $(\tfrac12, 1)$. By symmetry $\int f^+ \,d\lambda = \int f^- \,d\lambda = 1/\pi = 0.3183$, so $\int f\,d\lambda = 0$ and $\int |f|\,d\lambda = 2/\pi = 0.6366$.

The finiteness requirement is not decoration. Without it, the difference $\int f^+ - \int f^-$ can read $\infty - \infty$. On $\mathbb{R}$ with length, let $f = \mathbf{1}_{[0, \infty)} - \mathbf{1}_{(-\infty, 0)}$, equal to $1$ to the right of 0 and $-1$ to the left. Both $\int f^+$ and $\int f^-$ equal $\infty$. Adding up $f$ over $[-N, N]$ gives $0$, but adding it up over $[-N, 2N]$ gives $N$. Different ways of exhausting the line produce different answers, so there is no reason to prefer one of them, and the integral is simply not defined. Integrability rules out exactly this: when $\int |f| < \infty$, every sensible way of exhausting the space produces the same total.

Properties of the integral

The integral obeys four rules. Each is proved first for simple functions, then for non-negative functions through staircases, and finally for integrable functions by splitting into positive and negative parts; here only the meaning and the reasons are given.

• Linearity. For integrable $f, g$ and real $a, b$: $\int (af + bg)\,d\mu = a\int f\,d\mu + b\int g\,d\mu$. On $[0, 1]$, $\int x\,d\lambda = \tfrac12$ and $\int 4x(1 - x)\,d\lambda = \tfrac23$, so $\int \big(x + 4x(1 - x)\big)\,d\lambda = \tfrac76$.
• Monotonicity. If $f \le g$, then $\int f\,d\mu \le \int g\,d\mu$. Applied to $-|f| \le f \le |f|$ this gives $\big|\int f\,d\mu\big| \le \int |f|\,d\mu$: cancellations between positive and negative parts can only shrink a total.
• Null sets do not matter. If $f = g$ almost everywhere, then $\int f\,d\mu = \int g\,d\mu$. Let $N$ be the null set where they differ. Every simple function below $|f - g|$ vanishes outside $N$, so each of its terms $a_k \mu(A_k \cap N)$ is 0, and $\int |f - g|\,d\mu = 0$. Changing the hump $4x(1 - x)$ to the value 5 at every rational point leaves its integral at $2/3$, and $\int \mathbf{1}_{\mathbb{Q}}\,d\lambda = 0$.
• Zero integral. For $f \ge 0$, $\int f\,d\mu = 0$ exactly when $f = 0$ almost everywhere. If $f > 1/n$ on a set of measure $m$, then $f \ge \tfrac1n\mathbf{1}_{\{f > 1/n\}}$, and monotonicity gives $\int f\,d\mu \ge m/n$. So $\int f\,d\mu = 0$ forces every $\{f > 1/n\}$ to be null, and $\{f > 0\}$ is their countable union, which is null by "Lebesgue measure and null sets". The other direction is the previous rule.

The argument behind the last rule gives a general inequality: for $c > 0$, $c\cdot\mathbf{1}_{\{|f| > c\}} \le |f|$, hence

A function with a small integral can be large only on a small set. On the dart board, $\int s\,d\lambda = 2.0342$ and $c = 5$ give the bound $0.4068$ on the area where the score exceeds 5, while the true area, the centre disc of radius $0.1$, is $0.0314$. The bound is cheap and holds for every measure space. The probability part of this track meets it again under the name of Markov's inequality.

Integrals against other measures

Nothing in the construction used length. The integral was built from a measure, simple functions and a supremum, so it exists for every measure space $(\Omega, \mathcal{F}, \mu)$. The measures of "Measuring the size of sets" give familiar quantities, and each has a one-line formula.

• Counting measure on $\mathbb{N}$. A set $A$ has measure equal to its number of elements, so $\int \mathbf{1}_A\,d\mu = \sum_n \mathbf{1}_A(n)$, and extending through simple functions and staircases gives $\int f\,d\mu = \sum_{n=1}^{\infty} f(n)$. For $f(n) = 2^{-n}$ the integral is $1$, and for $f(n) = 1/n^2$ it is $\pi^2/6 = 1.6449$. On the six faces of a die with counting measure, $\int (\text{face})\,d\mu = 1 + 2 + \cdots + 6 = 21$.
• The Dirac measure $\delta_x$. All the mass sits at one point, so $\int f\,d\delta_x = f(x)$: integrating against $\delta_x$ means evaluating at $x$.
• Weighted point masses $\sum_i w_i \delta_{x_i}$. The integral is the weighted sum $\sum_i w_i f(x_i)$. For the masses $0.5$ at $0.2$, $0.3$ at $0.45$ and $0.2$ at $0.9$, and the function $f(x) = x$, it is $0.5 \cdot 0.2 + 0.3 \cdot 0.45 + 0.2 \cdot 0.9 = 0.415$.
• A measure with a density $\rho$. If $\mu(A) = \int_A \rho\,d\lambda$, then $\int f\,d\mu = \int f\rho\,d\lambda$. For an indicator this is the definition of $\mu$, linearity carries it to simple functions, and staircases carry it to all non-negative $f$. On the rod of "Measuring the size of sets", with $\rho(x) = 2x$, the integral of the position is $\int x\,d\mu = \int_0^1 x \cdot 2x\,dx = \tfrac23$.

The last integral has a physical meaning. The rod has total mass 1, and position times mass added over all its pieces is its balance point: $2/3$, right of the middle because the rod is heavier towards its right end. With length instead of the rod's mass, the same integral gives $\int x\,d\lambda = \tfrac12$.

One condition deserves care. On $\mathbb{N}$ with counting measure, $f$ is integrable exactly when $\sum_n |f(n)| < \infty$, that is, when the series converges absolutely. The alternating harmonic series $1 - \tfrac12 + \tfrac13 - \tfrac14 + \cdots$ converges, to $\ln 2 = 0.6931$, but $\sum 1/n = \infty$, so $f(n) = (-1)^{n+1}/n$ is not integrable. This is no accident. A measure space has no order on its points, so an integral cannot depend on the order in which they are visited, while the sum of a series that does not converge absolutely changes when its terms are rearranged: by a theorem of Riemann, it can be made to converge to any number. Absolute integrability is what makes the total independent of order.

Riemann and Lebesgue compared

Whenever a bounded function on an interval has a Riemann integral, it has a Lebesgue integral with the same value. The reason is short. The upper and lower sums of a Riemann partition are integrals of step functions, and step functions are simple functions. A lower step function lies below $f$ and an upper one above it, so the Lebesgue integral of $f$ lies between the lower and upper sums, and a Riemann integral exists exactly when those can be squeezed together.

The Lebesgue integral exists for more functions. The indicator $\mathbf{1}_{\mathbb{Q}}$ has no Riemann integral, but its Lebesgue integral is $1 \cdot \lambda(\mathbb{Q} \cap [0, 1]) = 0$. The two integrals also treat unbounded domains differently. The improper Riemann integral $\int_0^\infty \frac{\sin x}{x}\,dx$ is defined as the limit of $\int_0^b \frac{\sin x}{x}\,dx$ as $b \to \infty$, and the limit exists and equals $\pi/2 = 1.5708$: the first hump above the axis has area $1.8519$, and the following humps alternate in sign and shrink, so they cancel in a controlled way. The Lebesgue integral refuses to treat it as integrable. On the hump over $[j\pi, (j+1)\pi]$, $\int |\sin x|\,dx = 2$ and $1/x \ge 1/((j+1)\pi)$, so that hump of $|\sin x|/x$ has area at least $2/((j+1)\pi)$. These bounds form a harmonic series, which grows without limit: $10$ humps give at least $1.8646$, and it takes about $3.7$ million humps to pass $10$. So $\int_0^\infty |\sin x|/x\,dx = \infty$, and $\sin x / x$ is not integrable. This is the same $\infty - \infty$ problem as before: the improper Riemann integral depends on cancelling in a fixed order.

The last difference matters most for the next lesson. Riemann integrals do not survive limits. Enumerate the rationals as $q_1, q_2, \ldots$ and let $f_n$ be the indicator of the first $n$ of them. Each $f_n$ is zero except at finitely many points, so it has a Riemann integral, $0$. Yet the pointwise limit of the $f_n$ is $\mathbf{1}_{\mathbb{Q}}$, which has no Riemann integral at all. For the Lebesgue integral the limit is again integrable, and equals $0$. Under which conditions limits and integrals may be exchanged is the subject of "Convergence theorems".

Summary

• The Riemann integral slices the domain into columns and needs $f$ to be nearly constant on short stretches; the Lebesgue integral slices the range into layers and needs only the measures of level sets, as in sorting coins by value before counting them.
• For a simple function, $\int \varphi\,d\mu = \sum a_k \mu(A_k)$; for $f \ge 0$, $\int f\,d\mu$ is the supremum of the integrals of simple functions below $f$, approached by the dyadic staircases.
• The layer-cake formula $\int f\,d\mu = \int_0^\infty \mu(f > t)\,dt$ turns an integral over the space into an area under the tail function; on the dart board both routes give $2.0342$.
• A function is integrable when $\int |f|\,d\mu < \infty$, and then $\int f = \int f^+ - \int f^-$; the integral is linear and monotone, ignores null sets, and vanishes for $|f|$ only when $f = 0$ almost everywhere.
• Against counting measure the integral is a series, against $\delta_x$ an evaluation, against point masses a weighted sum, against a density $\rho$ the integral of $f\rho$; absolute convergence is what integrability means for series.
• The Lebesgue integral agrees with Riemann's wherever that exists, covers $\mathbf{1}_{\mathbb{Q}}$ and any measure space, and is the integral that behaves well under limits.