Egorov and Lusin convert measurable behavior into nearly uniform or nearly continuous behavior after discarding a set of small measure.
Egorov theorem
On finite measure spaces, almost-everywhere convergence is almost uniform convergence.
Theorem. If \(\mu(X)\lt \infty\) and \(f_n\to f\) almost everywhere, then for every \(\varepsilon>0\) there is \(E\subset X\) with \(\mu(E)\lt \varepsilon\) such that \(f_n\to f\) uniformly on \(X\setminus E\).
Proof. For \(k,m\), let \(A_{k,m}=\bigcup_{n\ge m}\{|f_n-f|>1/k\}\). For each \(k\), \(A_{k,m}\downarrow\varnothing\) up to a null set, so choose \(m_k\) with \(\mu(A_{k,m_k})\lt \varepsilon2^{-k}\). Let \(E=\bigcup_kA_{k,m_k}\). Outside \(E\), for every \(k\), all \(n\ge m_k\) satisfy \(|f_n-f|\le1/k\), which is uniform convergence.
Lusin theorem
Measurable functions are continuous outside small exceptional sets on Radon spaces.
Theorem. If \(f\) is finite-valued and measurable on a finite-measure Borel subset of \(\mathbb R^d\), then for every \(\varepsilon>0\) there is a compact \(K\) with \(\mu(X\setminus K)\lt \varepsilon\) such that \(f|_K\) is continuous.
Proof. Choose simple functions \(s_n\to f\) pointwise. Egorov gives a measurable \(E\) of measure \(<\varepsilon/2\) such that \(s_n\to f\) uniformly on \(X\setminus E\). Write each \(s_n\) using finitely many disjoint measurable level sets. By inner regularity, choose compact subsets of those level sets whose omitted measure, summed over all levels and all \(n\), is \(<\varepsilon/2\). Let \(K\) be the intersection of \(X\setminus E\) with the union of the retained compact pieces, followed by one final inner compact approximation if needed.
For fixed \(n\), the retained level-set pieces are finitely many disjoint compact sets and hence have positive pairwise distance; therefore \(s_n|_K\) is continuous. Since \(s_n\to f\) uniformly on \(K\), the uniform limit theorem makes \(f|_K\) continuous. The discarded set has measure below \(\varepsilon\).
Almost uniform convergence
Almost uniform convergence sits between almost-everywhere and uniform convergence.
Definition. \(f_n\to f\) almost uniformly if for every \(\varepsilon>0\) there is \(E\) with \(\mu(E)\lt \varepsilon\) such that convergence is uniform on \(X\setminus E\).
Proposition. Almost uniform convergence implies almost-everywhere convergence.
Proof. For each \(j\), choose \(E_j\) with \(\mu(E_j)<2^{-j}\) and uniform convergence on \(X\setminus E_j\). Subadditivity gives \[ \mu\!\left(\limsup_jE_j\right) \le\lim_{m\to\infty}\sum_{j\ge m}2^{-j}=0. \] If \(x\notin\limsup_jE_j\), then \(x\notin E_j\) for all sufficiently large \(j\). Choosing any one such \(j\), uniform convergence on \(X\setminus E_j\) implies \(f_n(x)\to f(x)\).
Measurable approximation by simple functions
Simple functions are dense in measure and in \(L^p\).
Theorem. If \(f\) is measurable and finite almost everywhere, then there are simple functions \(s_n\) with \(s_n\to f\) almost everywhere. If \(f\in L^p\), one can choose \(s_n\to f\) in \(L^p\).
Proof. Use dyadic quantization of truncated functions. Pointwise convergence follows as the truncation level and dyadic resolution increase. For \(L^p\), the errors are dominated by \(2^p|f|^p\) after truncation, and dominated convergence gives norm convergence.
Extension from indicators to measurable functions
Each approximation theorem identifies the topology in which a tractable class is dense.
Remark. The extension route must match the statement: indicators \(\to\) nonnegative simple functions uses linearity; simple \(\uparrow\) measurable uses monotone convergence; truncation plus domination reaches \(L^1\); \(L^p\)-density reaches bounded operators on \(L^p\). Naming that route prevents an approximation argument from silently changing its topology.