The Lebesgue integral is built from simple functions and monotone limits. Its strength is that limiting theorems are built into the construction rather than added afterward.
Simple functions
Simple functions are finite measurable step functions.
Definition. For \(s=\sum_{k=1}^m a_k\mathbf 1_{A_k}\) with \(a_k\ge0\) and \(A_k\) disjoint measurable sets, define \(\int s\,\mathrm{d}\mu=\sum_{k=1}^m a_k\mu(A_k)\).
Proposition. The integral of a nonnegative simple function is independent of the chosen disjoint representation.
Proof. Given two representations, refine them by all intersections \(A_i\cap B_j\). On this common partition the function has the same value cell by cell, and finite additivity of \(\mu\) makes both sums equal to the refined sum.
Proposition. For nonnegative simple \(s,t\) and \(a,b\ge0\), \[ \int(as+bt)\,\mathrm{d}\mu =a\int s\,\mathrm{d}\mu+b\int t\,\mathrm{d}\mu, \] and \(s\le t\) implies \(\int s\le\int t\).
Proof. Refine the level-set partitions of \(s\) and \(t\) by their pairwise intersections. On each common cell all functions are constant, so both assertions reduce to distributivity and monotonicity of finite sums with nonnegative coefficients.
Nonnegative measurable functions
General nonnegative integrals are suprema of simple lower approximations.
Definition. For measurable \(f\ge0\), define \[ \int f\,\mathrm{d}\mu=\sup\left\{\int s\,\mathrm{d}\mu:0\le s\le f,\ s\text{ simple}\right\}. \]
Theorem. If \(0\le f_n\uparrow f\), then \(\int f_n\,\mathrm{d}\mu\uparrow\int f\,\mathrm{d}\mu\).
Proof. Monotonicity gives the limit \(L\le\int f\). For a simple \(s\le f\) and \(0\lt c\lt 1\), the sets \(A_n=\{f_n\ge cs\}\) increase to \(X\) on \(\{s>0\}\). Thus \(\int f_n\ge c\int_{A_n}s\to c\int s\). Taking \(n\to\infty\), then \(c\uparrow1\), and then the supremum over \(s\), gives \(L\ge\int f\).
Corollary (countable additivity under the integral). If \(f_n\ge0\), then \[ \int\sum_{n=1}^{\infty}f_n\,\mathrm{d}\mu =\sum_{n=1}^{\infty}\int f_n\,\mathrm{d}\mu, \] with \(+\infty\) allowed.
Proof. The partial sums increase pointwise to \(\sum_nf_n\). Apply monotone convergence and finite additivity of the integral.
Integrable functions
Signed and complex functions are integrable when their absolute value is integrable.
Definition. A measurable real or complex function \(f\) is integrable if \(\int |f|\,\mathrm{d}\mu\lt \infty\). For real \(f\), write \(f=f^+-f^-\), and define \(\int f=\int f^+-\int f^-\).
Proposition. The integral is linear on \(L^1\), monotone on real functions, and satisfies \(|\int f\,\mathrm{d}\mu|\le\int|f|\,\mathrm{d}\mu\).
Proof. Choose simple \(s_n\uparrow f\) and \(t_n\uparrow g\) for nonnegative \(f,g\). Then \(as_n+bt_n\uparrow af+bg\), so monotone convergence and simple-function linearity prove nonnegative linearity. Decomposing signed functions into positive and negative parts extends it to \(L^1\). Monotonicity is immediate from the defining supremum. For real \(f\), \(-|f|\le f\le|f|\) gives the absolute-value inequality. For complex \(f\), choose \(\theta\) such that \(e^{-i\theta}\int f=|\int f|\); then \[ \left|\int f\right| =\operatorname{Re}\int e^{-i\theta}f \le\int|f|. \]
Proposition (zero integral). If \(f\ge0\), then \(\int f\,\mathrm{d}\mu=0\) if and only if \(f=0\) almost everywhere.
Proof. If the integral is zero, then for every \(n\), \[ 0=\int f\ge\frac1n\mu(f\ge1/n), \] so each set \(\{f\ge1/n\}\) is null. Their union is \(\{f>0\}\). The converse follows because changing a nonnegative measurable function on a null set does not change any simple lower approximation's integral.
Layer-cake representation
Nonnegative integrals can be computed from tail measures.
Theorem. If \(f\ge0\), then \[ \int f\,\mathrm{d}\mu=\int_0^\infty \mu(f>t)\,\mathrm{d}t. \]
Proof. For a simple function \(s=\sum_ka_k\mathbf1_{A_k}\), direct integration of the decreasing step function \(t\mapsto\mu(s>t)\) gives \[ \int_0^\infty\mu(s>t)\,\mathrm{d}t =\sum_ka_k\mu(A_k)=\int s\,\mathrm{d}\mu. \] Choose simple \(s_n\uparrow f\). Then \(\mu(s_n>t)\uparrow\mu(f>t)\) for every \(t\) except possibly the countable set of levels at which the approximation convention hits equality. Applying monotone convergence first to \(s_n\) on \(X\) and then to the tail functions on \([0,\infty)\) yields the formula. This proof does not require product measure; after Tonelli is established, the identity is also the interchange of the two integrals in \(f(x)=\int_0^\infty\mathbf1_{\{t<f(x)\}}\,\mathrm{d}t\).
Corollary (tail moments). For \(p>0\) and measurable \(f\), \[ \int |f|^p\,\mathrm{d}\mu =p\int_0^\infty t^{p-1}\mu(|f|>t)\,\mathrm{d}t. \]
Proof. Apply layer-cake to \(|f|^p\) and substitute \(s=t^p\), so \(\mathrm{d}s=pt^{p-1}\,\mathrm{d}t\).
Corollary (Markov inequality). For \(f\ge0\) and \(t>0\), \[ \mu(f\ge t)\le\frac1t\int f\,\mathrm{d}\mu. \]
Proof. Since \(f\ge t\mathbf1_{\{f\ge t\}}\), integrate both sides.
Absolute continuity of the integral
Integrable functions have uniformly small integral over small-measure sets.
Proposition. If \(f\in L^1(\mu)\), then for every \(\varepsilon>0\) there is \(\delta>0\) such that \(\mu(A)\lt \delta\Rightarrow\int_A |f|\,\mathrm{d}\mu\lt \varepsilon\).
Proof. Choose \(M\) so large that \(\int_{\{|f|>M\}}|f|\,\mathrm{d}\mu\lt \varepsilon/2\). Then \(\int_A|f|\le M\mu(A)+\varepsilon/2\). Take \(\delta=\varepsilon/(2M)\), with the trivial modification if \(M=0\).