Measurable functions are functions compatible with sigma-algebras. They are the objects that can be integrated, approximated by simple functions, and compared almost everywhere.
Measurability criteria
For real-valued functions it is enough to test half-lines.
Theorem. A function \(f:X\to\mathbb R\) is measurable if and only if \(\{x:f(x)>a\}\in\mathcal A\) for every \(a\in\mathbb R\). Equivalently one may use \(\{f\le a\}\), \(\{f\lt a\}\), or \(\{f\ge a\}\).
Proof. The inverse images of intervals \((a,\infty)\) generate the Borel sigma-algebra on \(\mathbb R\), so measurability implies the condition. Conversely, if inverse images of these generators are measurable, the class of Borel sets with measurable inverse image is a sigma-algebra containing the generators.
Proposition (measurable calculus). If \(f,g:X\to\overline{\mathbb R}\) are measurable, then \(f+g\), \(fg\), \(\max(f,g)\), and \(\min(f,g)\) are measurable wherever the extended-real operations are defined. If \(f_n\) are measurable, then \[ \sup_nf_n,\quad\inf_nf_n,\quad\limsup_nf_n,\quad\liminf_nf_n \] are measurable.
Proof. The map \(x\mapsto(f(x),g(x))\) into \(\mathbb R^2\) is measurable because inverse images of measurable rectangles are intersections of measurable sets. Composition with a continuous arithmetic map is measurable. For suprema, \[ \left\{\sup_nf_n>a\right\}=\bigcup_n\{f_n>a\}, \] and infima follow by negation. Finally, \[ \limsup_nf_n=\inf_N\sup_{n\ge N}f_n,\qquad \liminf_nf_n=\sup_N\inf_{n\ge N}f_n. \]
Simple approximation
Nonnegative measurable functions are monotone limits of simple functions.
Theorem. If \(f\ge0\) is measurable, there exist nonnegative simple functions \(s_n\uparrow f\) pointwise.
Proof. Define \(s_n(x)=2^{-n}\lfloor 2^n\min(f(x),n)\rfloor\). Each \(s_n\) takes finitely many values and is measurable because its level sets are described by inverse images of intervals under \(f\). The construction increases to \(f\) pointwise.
Corollary. Every real measurable function is a pointwise limit of finite-valued measurable functions. If it is bounded, the approximation can be uniform.
Proof. Apply the nonnegative construction to \(f^+\) and \(f^-\), then subtract. If \(|f|\le M\), dyadic rounding without truncation gives simple \(s_n\) with \(|s_n-f|\le2^{-n}\) everywhere.
Almost-everywhere equality
Measure theory identifies functions that differ only on null sets.
Definition. Functions \(f,g\) are equal almost everywhere if \(\mu(\{f\ne g\})=0\). A property holds almost everywhere if it fails only on a null set.
Proposition. If \(f=g\) almost everywhere and \(f\) is measurable on a complete measure space, then \(g\) is measurable.
Proof. For every \(a\), the symmetric difference between \(\{g>a\}\) and \(\{f>a\}\) is contained in \(\{f\ne g\}\). Completeness makes the exceptional subset measurable, so \(\{g>a\}\) is measurable.
Convergence in measure
Convergence in measure ignores small exceptional sets whose measure vanishes.
Definition. \(f_n\to f\) in measure if for every \(\varepsilon>0\), \(\mu(|f_n-f|>\varepsilon)\to0\).
Theorem. If \(f_n\to f\) almost everywhere and \(\mu(X)\lt \infty\), then \(f_n\to f\) in measure.
Proof. For fixed \(\varepsilon>0\), the sets \(A_n=\bigcup_{k\ge n}\{|f_k-f|>\varepsilon\}\) decrease to a null set by almost-everywhere convergence. Since \(\mu(X)\lt \infty\), continuity from above gives \(\mu(A_n)\to0\). Then \(\mu(|f_n-f|>\varepsilon)\le\mu(A_n)\).
Example (finite measure is necessary). On \((\mathbb R,m)\), let \(f_n=\mathbf1_{[n,n+1]}\). Then \(f_n(x)\to0\) for every \(x\), but \[ m(|f_n|>1/2)=1 \] for every \(n\), so convergence in measure fails.
Proposition (uniqueness). If \(f_n\to f\) in measure and \(f_n\to g\) in measure, then \(f=g\) almost everywhere.
Proof. For \(\varepsilon>0\), \[ \{|f-g|>\varepsilon\} \subset \{|f-f_n|>\varepsilon/2\}\cup \{|f_n-g|>\varepsilon/2\}. \] The measure of the right side tends to zero, so \(\mu(|f-g|>\varepsilon)=0\). Taking the union over positive rational \(\varepsilon\) gives \(f=g\) almost everywhere.
Subsequence principle
Convergence in measure contains almost-everywhere convergence along subsequences.
Theorem. If \(f_n\to f\) in measure, then some subsequence converges to \(f\) almost everywhere.
Proof. Choose \(n_k\) so that \[ \mu(|f_{n_k}-f|>2^{-k})<2^{-k}. \] Let \(E_k\) denote this exceptional set. Since \(\sum_k\mu(E_k)<\infty\), \[ \mu\!\left(\limsup_kE_k\right) =\mu\!\left(\bigcap_m\bigcup_{k\ge m}E_k\right) \le\lim_{m\to\infty}\sum_{k\ge m}\mu(E_k)=0. \] Outside this null set, \(|f_{n_k}(x)-f(x)|\le2^{-k}\) eventually, proving almost-everywhere convergence. This is the first Borel-Cantelli argument, derived directly from subadditivity.