# Another version of Egoroff's theorem

Let $(X,M,\mu)$ be a finite or $\sigma$-finite measure space. Let $\{f_n\}$ be a sequence of finite a.e., measurable functions such that $f_n \to f$ [a.e.]. Then there is a partition of $X$ into a sequence $E_0,\,E_1,\,E_2,\,\dots$ of disjoint measurable sets such that $\mu(E_0) = 0$ and $f_n \to f$ uniformly on each $E_i$, $i \geqslant 1$.

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By the way, welcome to math.stackexchange. – Davide Giraudo Nov 12 '12 at 21:00

Let $\{A_k\}\subset\cal M$ pairwise disjoint of finite measure such that $X=\bigcup_{k\in\Bbb N}A_k$. For each integers $j,k$, using Egoroff theorem, we can find $S_{k,j}\subset A_k$ such that $\mu(A_k\setminus S_{k,j})\leqslant 2^{-k-j}$ and $\lim_{n\to +\infty}\sup_{x\in S_{k,j}}|f_n(x)-f(x)|\to 0$.
Let $E_0:=\bigcap_{j\geqslant 1}\bigcup_{k\geqslant 1}(A_k\setminus S_{k,j})$. We assume the sequences $\{S_{k,j}\}_j$ increasing. let $T_{k,j}:=S_{k,j}\setminus S_{k,j-1}$· Then the sets $T_{k,j}$ are pairwise disjoint and we have uniform convergence on these sets.