The Lebesgue Integral

With measure theory in hand we can define an integral that repairs the two structural defects of the Riemann integral (riemann-integral.md §7): its integrable class is too small (the Dirichlet function is excluded), and it does not commute with pointwise limits. The Lebesgue integral fixes both, at the price of the measure-theoretic apparatus, and it is the integral used everywhere downstream — the $L^p$ spaces of functional analysis, the expectation of probability, and the densities that Radon–Nikodym produces in §7.

Throughout, $(X,\mathcal{A},\mu )$ is a measure space and "measurable" means $\mathcal{A}$-measurable.

References: Rudin, Real and Complex Analysis, ch. 1–3, 6; Folland, Real Analysis, ch. 2–3.

1. Slice the Range, Not the Domain

Riemann partitions the domain and asks how much $f$ varies over each piece — which is why wild oscillation (the Dirichlet function) is fatal. Lebesgue partitions the range: for each layer $[c,c+\Delta c)$ ask how big is the set where $f$ takes values in that layer, and add $c\cdot \mu (\text{that set})$. Wild oscillation costs nothing; the only requirement is that the level sets be measurable, which is precisely the definition of a measurable function (measure-theory.md §7).

The construction proceeds in three steps — simple $\to$ non-negative $\to$ general — each defined so that the previous one is a special case.

2. Simple Functions

A simple function is a measurable $s:X\to [0,\infty )$ with finite range; writing its distinct values $c_1,\dots ,c_n$ and $A_i=s^{-1}(c_i)$ gives the canonical form $s={\sum }_{i=1}^n{c}_i{1}_{A_i}$ with the $A_i$ disjoint and measurable.

Definition. ${\int }_{X}sd\mu :=\sum _{i=1}^n{c}_i\mu (A_i)>\in [0,\infty ]$, with the convention $0\cdot \infty =0$.

The convention is what makes $\int 0d\mu =0$ on a space of infinite measure, and it is forced if the theory is to be consistent. One checks directly that the value is independent of the representation, that $\int$ is additive and positively homogeneous on simple functions, and that $s\le t⟹\int s\le \int t$.

3. Non-Negative Functions

Definition. For measurable $f:X\to [0,\infty ]$,

$${\int }_{X}fd\mu :=\sup \{{\int }_{X}sd\mu :s\text{ simple},0\le s\le f\}.$$

The supremum always exists in $[0,\infty ]$, so every non-negative measurable function has an integral — possibly $+\infty$. This is a real difference from the Riemann theory, where integrability is a nontrivial hypothesis; here the hypothesis reappears only when we ask for a finite value.

By simple approximation (measure-theory.md §7) there are simple $s_k\uparrow f$, and the next theorem says the sup is a limit along any such sequence.

4. The Convergence Theorems

These three results are the reason the Lebesgue integral exists. All are statements about exchanging $\lim$ and $\int$, and all fail for the Riemann integral.

Monotone Convergence Theorem (MCT). If $0\le f_1\le f_2\le \cdots$ are measurable and $f_k\uparrow f$ pointwise, then $\int f_{k}d\mu \uparrow \int fd\mu$.

Proof sketch. "$\le$" is monotonicity. For "$\ge$", fix simple $s\le f$ and $c\in (0,1)$, and set $E_k=\{f_k\ge cs\}$. Then $E_k\uparrow X$, and continuity of $\mu$ from below (measure-theory.md §3) gives $\int f_k\ge c{\int }_{E_k}s\to c\int s$. Let $c\uparrow 1$ and take the sup over $s$. $\square$

Note where the hypotheses are consumed: countable additivity enters exactly once, as continuity from below. MCT is the countable additivity of $\mu$ transposed to functions — which is why finitely additive measures have no useful integration theory (finite-additivity.md).

Corollary (countable additivity of the integral). For measurable $f_k\ge 0$, $\int {\sum }_k{f}_{k}d\mu ={\sum }_k\int f_{k}d\mu$. Sums and integrals of non-negative terms always commute.

Fatou's Lemma. For measurable $f_k\ge 0$, $\int \underset{k}{\mathrm{liminf}}{f}_{k}d\mu \le \underset{k}{\mathrm{liminf}}\int f_{k}d\mu .$

Proof. Apply MCT to $g_k={\inf }_{j\ge k}{f}_j\uparrow \mathrm{liminf}{f}_k$, noting $g_k\le f_j$ for $j\ge k$. $\square$

The inequality can be strict: the "escaping mass" examples $f_k=k{1}_{(0,1/k)}$ (mass concentrates) and $f_k=1_{[k,k+1]}$ (mass escapes to infinity) both have $\mathrm{liminf}{f}_k=0$ pointwise but $\int f_k=1$. Fatou is the correct one-sided statement in general; equality needs a dominating hypothesis.

Dominated Convergence Theorem (DCT). If $f_k\to f$ pointwise a.e. and there is a dominating $g\in L^1(\mu )$ (i.e. $\int gd\mu <\infty$) with $|f_k|\le g$ a.e. for all $k$, then $f\in L^1$ and

$$\underset{k\to \infty }{\lim }{\int }_X{f}_{k}d\mu ={\int }_{X}fd\mu ,\text{indeed}\int |f_k-f|d\mu \to 0.$$

Proof. Apply Fatou to $2g-|f_k-f|\ge 0$. $\square$

Both escaping-mass examples above violate domination (any $g$ would need $\int g=\infty$), which is exactly what the theorem asks for. DCT is the workhorse: differentiating under the integral sign, continuity of Fourier transforms, and the completeness of $L^p$ all reduce to it.

5. General and Complex-Valued Functions

For measurable $f:X\to \overline{\mathbb{R}}$ write $f=f^{+}-f^{-}$ with $f^{\pm }=\max (\pm f,0)\ge 0$.

Definition. $f$ is integrable ($f\in L^1(\mu )$) if $\int f^{+},\int f^{-}><\infty$, equivalently $\int |f|d\mu <\infty$; then $\int fd\mu =>\int f^{+}d\mu -\int f^{-}d\mu$. Complex $f$ is integrable iff $|f|$ is, with $\int f=\int \mathrm{Re}f+i\int \mathrm{Im}f$.

The Lebesgue integral is absolute: $f$ integrable $⟺$ $|f|$ integrable. Conditionally convergent objects such as ${\int }_0^{\infty }\frac{\sin x}{x}dx$ are therefore not Lebesgue integrals but improper limits — a genuine expressive loss relative to the Riemann/gauge theories, and the reason improper integrals keep their own notation in physics texts.

Basic properties: linearity on $L^1$, monotonicity, the triangle inequality $|\int f|\le \int |f|$, and

Proposition. ${\int }_X|f|d\mu =0⟺f=0$ a.e. Consequently the integral does not distinguish functions equal a.e.

6. Comparison with Riemann

Theorem. If $f$ is Riemann integrable on $[a,b]$ then it is Lebesgue integrable with respect to ${\lambda }^1$ and the two integrals agree.

So nothing computed in riemann-integral.md or multiple-integrals.md changes value; the Fundamental Theorem of Calculus remains the practical evaluation tool.

Theorem (Lebesgue's criterion). A bounded $f:[a,b]\to \mathbb{R}$ is Riemann integrable iff its set of discontinuities is ${\lambda }^1$-null.

Proof sketch. Define the oscillation ${\mathrm{osc}}_f(x)={\lim }_{\delta \to 0}\mathrm{diam}f(B_{\delta }(x))$; $f$ is continuous at $x$ iff ${\mathrm{osc}}_f(x)=0$, and $D_{\epsilon }=\{{\mathrm{osc}}_f\ge \epsilon \}$ is compact. If the discontinuity set $D={\bigcup }_k{D}_{1/k}$ is null, each $D_{\epsilon }$ is null and compact, hence coverable by finitely many intervals of small total length; on the complement the oscillation is $<\epsilon$, and splitting a partition accordingly gives $U(f,P)-L(f,P)<\epsilon (b-a+\mathrm{osc})$, the Riemann criterion. Conversely, if ${\lambda }^1(D_{\epsilon })>0$ for some $\epsilon$, every partition has $U-L\ge \epsilon {\lambda }^1(D_{\epsilon })>0$. $\square$

This discharges the IOU left open in riemann-integral.md §3 and multiple-integrals.md §2. The Dirichlet function $1_{\mathbb{Q}}$ is discontinuous everywhere, hence not Riemann integrable, but as a function vanishing a.e. it has Lebesgue integral $0$ — the standard illustration that the Lebesgue class is strictly larger.

7. Densities: Radon–Nikodym

Integration produces new measures from old, and the converse question — which measures arise this way — is answered completely.

Definition. $\nu$ is absolutely continuous with respect to $\mu$ ($\nu \ll \mu$) if $\mu (A)=0⟹\nu (A)=0$. It is singular ($\nu \perp \mu$) if the two are carried by disjoint sets.

Theorem (Radon–Nikodym). If $\mu ,\nu$ are $\sigma$-finite and $\nu \ll \mu$, there is a measurable $h\ge 0$, unique up to $\mu$-a.e. equality, with

$$\nu (A)={\int }_{A}hd\mu \text{for all }A\in \mathcal{A}.$$

$h=\frac{d\nu }{d\mu }$ is the Radon–Nikodym derivative.

Theorem (Lebesgue decomposition). Any $\sigma$-finite $\nu$ splits uniquely as $\nu ={\nu }_{ac}+{\nu }_s$ with ${\nu }_{ac}\ll \mu$ and ${\nu }_s\perp \mu$.

The slick proof of Radon–Nikodym (von Neumann's) runs through the Riesz representation theorem in the Hilbert space $L^2(\mu +\nu )$ — see hilbert-spaces.md. Two downstream uses fix its importance here:

• Probability densities. A distribution on $\mathbb{R}$ has a density $p$ exactly when it is $\ll {\lambda }^1$, and then $p=d\nu /d{\lambda }^1$ (random-variables.md). Distributions with atoms, or on the Cantor set, are the singular part.
• Conditional expectation. $E[X\mid \mathcal{G}]$ is defined as a Radon–Nikodym derivative of a signed measure restricted to $\mathcal{G}$ (conditional-expectation.md).

8. $L^p$ Spaces

Definition. For $1\le p<\infty$, $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$ and $L^p(\mu )$ is the set of measurable $f$ with $\Vert f{\Vert }_p<\infty$, modulo equality a.e. For $p=\infty$, $\Vert f{\Vert }_{\infty }$ is the essential supremum.

Quotienting by a.e. equality is not fastidiousness — without it $\Vert \cdot {\Vert }_p$ is only a seminorm (§5's proposition), and the space is not normed.

Hölder. $\Vert fg{\Vert }_1\le \Vert f{\Vert }_p\Vert g{\Vert }_q$ for $\frac{1}{p}+\frac{1}{q}=1$. Minkowski. $\Vert f+g{\Vert }_p\le \Vert f{\Vert }_p+\Vert g{\Vert }_p$ — the triangle inequality. Riesz–Fischer. $L^p(\mu )$ is complete: a Banach space, and $L^2$ a Hilbert space.

Proof of completeness (sketch). Given a Cauchy sequence, extract a subsequence with $\Vert f_{k+1}-f_k{\Vert }_p\le 2^{-k}$; the series $\sum |f_{k+1}-f_k|$ has finite $L^p$ norm by MCT and Minkowski, hence converges a.e.; DCT identifies the limit and gives convergence in norm. $\square$

Completeness is exactly the property that functional analysis assumes when it lists $L^p$ among the standard Banach spaces, and the fact that $L^2$ is a Hilbert space is what makes it the state space of quantum mechanics.

Convergence in $L^p$ vs. a.e. Neither implies the other: the "typewriter" sequence of indicators of $[j/2^k,(j+1)/2^k]$ tends to $0$ in $L^1$ but nowhere pointwise, while $k{1}_{(0,1/k)}\to 0$ pointwise but not in $L^1$. What is true: $L^p$ convergence implies a.e. convergence along a subsequence — the extraction step in the proof above. The systematic comparison of convergence modes is convergence-modes.md.

9. Where This Is Used

• Product measures and Fubini — iterated integration.
• Expectation — $E[X]={\int }_{\Omega }XdP$; MCT/Fatou/DCT reappear verbatim as the convergence theorems of probability.
• Conditional expectation — Radon–Nikodym.
• Normed and Banach spaces, Hilbert spaces — $L^p$, $L^2$.