For Lebesgue measure \(m\) on \(\RR^d\) and \(\varepsilon>0\):
| Principle | Theorem |
|---|---|
| A measurable set \(E\) with \(m(E)<\infty\) differs from a finite union of rectangles by a set of measure \(<\varepsilon\) | regularity of Lebesgue measure |
| A measurable function finite a.e. on \(E\), \(m(E)<\infty\), is continuous on a closed \(F\subseteq E\) with \(m(E\setminus F)<\varepsilon\) | Lusin |
| A sequence converging a.e. on \(E\), \(m(E)<\infty\), to an a.e. finite limit converges uniformly on a closed \(A\subseteq E\) with \(m(E\setminus A)<\varepsilon\) | Egorov |
(Egorov's theorem and bounded convergence)
Egorov’s theorem proves the bounded convergence theorem. Let \(\abs{f_n}\leq M\) vanish outside \(E\), \(m(E)<\infty\), and \(f_n\to f\) almost everywhere. For \(\varepsilon>0\) take a closed \(A\subseteq E\) with \(m(E\setminus A)<\varepsilon\) on which \(f_n\to f\) uniformly. Then \(\int_E\abs{f_n-f}\leq m(E)\sup_A\abs{f_n-f}+2M\varepsilon\), so \(\limsup_n\int\abs{f_n-f}\leq2M\varepsilon\) for every \(\varepsilon>0\).
(Egorov's theorem needs finite measure)
On \(\RR\), \(f_n\coloneqq\chi_{[n,n+1]}\to0\) pointwise, but \(\sup_{A}\abs{f_n}=1\) for all \(n\) on any \(A\) with \(m(\RR\setminus A)<1\), since such \(A\) meets every \([n,n+1]\) in positive measure.
The statements and proofs are on The convergence theorems and Littlewood's principles: proofs.