Here is a statement of Egoroff's Theorem:
Egoroff's Theorem. Suppose that $\mu(X) < \infty$, and $f, f_1, f_2, f_3, \ldots$ are all measurable complex-valued functions on $X$ such that $f_n \to f$ almost everywhere. Then for every $\varepsilon > 0$, there exists $E \subseteq X$ such that $\mu(E) < \varepsilon$ and $f_n \to f$ uniformly on $E^c$ (the complement of $E$ in $X$).
Is it possible to replace the "$\mu(E) < \varepsilon$" condition by "$\mu(E) = 0$" in the statement and still have the theorem be true? I suspect that the answer is no, but I am having trouble thinking of an appropriate counterexample.