The Lebesgue Integral

The Lebesgue integral adds up a function by slicing its range — the output values — rather than its domain. A Riemann integral chops the input axis into thin vertical strips; the Lebesgue integral instead chops the value axis into bands and asks, for each band, how much of the space lands that high. Measured against a measure \mu, we write it

\int_{\Omega} f \, d\mu.

Slicing by value is exactly what lets the integral handle badly-behaved measurable functions and — crucially — pass cleanly through limits, the convergence theorems the rest of the theory leans on. It is built in four deliberate stages, each extending the last: first the indicator of a single set, then a finite combination of indicators (a simple function), then any non-negative measurable function, and finally a general signed one. We take them in order and skip no step.

Stage 0 — the indicator of a set

The atom of the whole construction is the indicator function of a measurable set A \in \mathcal{F}: the function that is 1 on A and 0 off it,

\mathbf{1}_A(\omega) \;=\; \begin{cases} 1 & \omega \in A,\\[2pt] 0 & \omega \notin A. \end{cases}

Its integral is defined to be the measure of the set it marks out — there is nothing to compute, this is the seed everything else grows from:

\int_{\Omega} \mathbf{1}_A \, d\mu \;=\; \mu(A).

Read it as: "weight 1 over the region A, weight 0 everywhere else, and total it up" — and the total weight is just the size \mu(A) of A. With \mu = \mathbb{P} a probability measure this already says \int \mathbf{1}_A \, d\mathbb{P} = \mathbb{P}(A), the fact that makes probability a special case of integration.

Stage 1 — simple functions

A simple function takes finitely many values a_1, \dots, a_n, each on a measurable set A_i, where the A_i partition \Omega (they are disjoint and cover everything). It is just a finite stack of indicators, one per value:

S \;=\; \sum_{i=1}^{n} a_i \, \mathbf{1}_{A_i}.

Its integral is forced on us by Stage 0 and the demand that integration be additive: integrate the stack term by term, and each term a_i \mathbf{1}_{A_i} contributes a_i \mu(A_i). So the integral is the value-weighted total of the measures of those sets,

\int_{\Omega} S \, d\mu \;=\; \sum_{i=1}^{n} a_i \, \mu(A_i).

In particular, taking n = 1 with S = \mathbf{1}_A recovers Stage 0, \int_{\Omega} \mathbf{1}_A \, d\mu = \mu(A).

Linearity on simple functions, line by line

The one fact the rest of the theory rests on is that S \mapsto \int S \, d\mu is linear. It is worth seeing exactly why, because the only subtlety is that two simple functions are written over different partitions, and we must put them on a common one before we can add.

Take two simple functions over their own partitions of \Omega,

S \;=\; \sum_{i=1}^{m} a_i \, \mathbf{1}_{A_i}, \qquad T \;=\; \sum_{j=1}^{n} b_j \, \mathbf{1}_{B_j}.

Form the common refinement: the sets C_{ij} = A_i \cap B_j. Because \{A_i\} and \{B_j\} each partition \Omega, the C_{ij} are disjoint and also partition \Omega; and crucially, for a fixed i the set A_i is sliced exactly into its pieces C_{ij}, so by finite additivity of the measure \mu,

\mu(A_i) \;=\; \sum_{j=1}^{n} \mu(C_{ij}), \qquad \mu(B_j) \;=\; \sum_{i=1}^{m} \mu(C_{ij}).

On each tiny cell C_{ij} both functions are constant — S = a_i and T = b_j there — so for any scalars \alpha, \beta the combination \alpha S + \beta T is itself simple, equal to \alpha a_i + \beta b_j on C_{ij}:

\alpha S + \beta T \;=\; \sum_{i=1}^{m}\sum_{j=1}^{n} (\alpha a_i + \beta b_j)\, \mathbf{1}_{C_{ij}}.

Now apply the Stage 1 definition to this simple function and unpack it step by step:

\int (\alpha S + \beta T)\, d\mu \;=\; \sum_{i=1}^{m}\sum_{j=1}^{n} (\alpha a_i + \beta b_j)\, \mu(C_{ij})

by the definition of the integral of a simple function. Split the sum across the plus sign:

=\; \alpha \sum_{i=1}^{m}\sum_{j=1}^{n} a_i\, \mu(C_{ij}) \;+\; \beta \sum_{i=1}^{m}\sum_{j=1}^{n} b_j\, \mu(C_{ij})

by regrouping the finite double sum. In the first double sum a_i does not depend on j, so sum over j first; in the second b_j does not depend on i, so sum over i first:

=\; \alpha \sum_{i=1}^{m} a_i \Big(\sum_{j=1}^{n}\mu(C_{ij})\Big) \;+\; \beta \sum_{j=1}^{n} b_j \Big(\sum_{i=1}^{m}\mu(C_{ij})\Big)

and the two bracketed sums are exactly the additivity identities above, so they collapse:

=\; \alpha \sum_{i=1}^{m} a_i\, \mu(A_i) \;+\; \beta \sum_{j=1}^{n} b_j\, \mu(B_j) \;=\; \alpha \int S \, d\mu \;+\; \beta \int T \, d\mu.

That is linearity, with no gaps: the common refinement is the whole trick, and additivity of \mu does the rest. Monotonicity is even quicker — if S \le T, then on every cell a_i \le b_j, and since each \mu(C_{ij}) \ge 0 the term-by-term inequality survives summation, giving \int S \, d\mu \le \int T \, d\mu.

Stage 2 — non-negative functions

For a general measurable f \ge 0 we cannot write down a finite formula, so instead we approximate it from below by simple functions and take the best such approximation — the supremum of their integrals:

\int_{\Omega} f \, d\mu \;=\; \sup\Big\{\, \textstyle\int_{\Omega} S \, d\mu \;:\; S \text{ simple},\; 0 \le S \le f \,\Big\}.

Two things make this definition sound. First, the set of candidates is non-empty — the simple function S \equiv 0 always qualifies — so the supremum is over a non-empty set of real numbers. Second, by monotonicity on simple functions (just proved), every candidate already satisfies \int S \, d\mu \le \int S' \, d\mu whenever S \le S', so raising the staircase can only raise the estimate — the supremum genuinely captures the "best from below".

The trick is to partition the range, not the domain: slice the y-axis into finer and finer levels and, on each level, ask "how much of the space reaches this high?". Concretely, with n levels of height h = \tfrac{1}{n} one can take the simple function that floors f down to the nearest level below it; as n grows the staircase climbs up to f and the simple-function integrals increase to \int f \, d\mu. The interactive figure below shows exactly this staircase tightening against the curve.

Let 0 \le f_1 \le f_2 \le \cdots be a non-decreasing sequence of non-negative measurable functions with pointwise limit f = \lim_{n\to\infty} f_n. Then the limit and the integral may be exchanged: \int_{\Omega} \Big(\lim_{n\to\infty} f_n\Big)\, d\mu \;=\; \lim_{n\to\infty} \int_{\Omega} f_n \, d\mu. This is precisely what licenses the Stage 2 definition: the flooring staircases S_n \uparrow f are such a non-decreasing sequence, so their integrals rise to \int f \, d\mu with no loss in the limit.

Stage 3 — general functions

A general f takes both signs, but Stage 2 only knows how to integrate non-negative functions. So split f into its positive and negative parts,

f^{+} \;=\; \max(f, 0), \qquad f^{-} \;=\; \max(-f, 0),

each of which is non-negative and measurable. They reconstruct f and its absolute value cleanly:

f \;=\; f^{+} - f^{-}, \qquad |f| \;=\; f^{+} + f^{-},

and at every point \omega exactly one of f^{+}(\omega), f^{-}(\omega) is non-zero — the positive and negative parts are disjointly supported. Each part is non-negative, so each has a Stage 2 integral; define

\int_{\Omega} f \, d\mu \;=\; \int_{\Omega} f^{+} \, d\mu \;-\; \int_{\Omega} f^{-} \, d\mu,

provided the two pieces are not both +\infty (otherwise the difference is the meaningless \infty - \infty). We call f integrable precisely when both pieces are finite, which by |f| = f^{+} + f^{-} and Stage 1 linearity is the single condition

\int_{\Omega} |f| \, d\mu \;=\; \int_{\Omega} f^{+} \, d\mu + \int_{\Omega} f^{-} \, d\mu \;<\; \infty.

Two properties now extend from the simple-function case to all integrable f, g — by approximating each part from below and passing to the limit with Monotone Convergence — and they are used on essentially every page that follows:

The picture that separates the two integrals is which axis you cut. The Riemann integral chops the domain into thin vertical strips and sums f(x_i)\,\Delta x_i — it asks "where is x?". The Lebesgue integral chops the range into horizontal bands and, for each band height y, weighs the set \{\,\omega : f(\omega) \ge y\,\} by \mu — it asks "how much of the space reaches this high?". Slicing the range is what frees the integral from needing a tidy domain: the indicator of the rationals, \mathbf{1}_{\mathbb{Q}}, has no Riemann integral (its strips never settle), yet its Lebesgue integral is plainly \mu(\mathbb{Q}) = 0.

Slicing the range is also why limits pass through so cleanly — the single most important advantage for probability, where we constantly take limits of approximating random variables and of conditional estimates. Monotone Convergence handles increasing sequences; its dominated cousin handles the general case:

Dominated Convergence. If f_n \to f pointwise and all the f_n are bounded by one fixed integrable g (so |f_n| \le g with \int g \, d\mu < \infty), then

\lim_{n\to\infty} \int_{\Omega} f_n \, d\mu \;=\; \int_{\Omega} \lim_{n\to\infty} f_n \, d\mu \;=\; \int_{\Omega} f \, d\mu.

The dominating g is the safety net that stops mass from leaking off to infinity. This is the theorem that lets us swap a limit and an expectation — exactly the move behind continuity of expectation, the tower property of conditioning, and the interchange of differentiation and integration that pricing formulas rely on.

The integral of an indicator is the measure of the set, not automatically 1: \int \mathbf{1}_A\,d\mu = \mu(A). And simple functions add by value-weighted measure, \int \sum_i a_i\mathbf{1}_{A_i}\,d\mu = \sum_i a_i\mu(A_i) — you cannot drop the measures or just sum the heights. Linearity is with respect to the integrand, not a licence to ignore the underlying measure.