Necessary condition in Egoroff's Theorem Construct a sequence $\{f_n\}$ of measurable functions on $[0,1]$ such that the sequence converges everywhere on $[0,1]$ but for any set $B\subset [0,1]$ of measure $1$, the sequence fails to converge uniformly on $B$..
This says it is not always possible to have $\mu(A\setminus B)=0$ in case of Egoroff's theorem.
First of all i need some sequence of functions that does not converge uniformly..
I take characteristic functions $f_n=\chi_{[0,\frac{1}{n}]}$ on $[0,1]$.. Given any $x<1$ there exists $N\in \mathbb{N}$ such that $\frac{1}{N}<x$ so, $f_N(x)=0$ and $f_n(x)=0$ for all $n\geq N$.. So, $f_n$ converges pointwise to constant function zero everywhere.. I am having trouble in showing that this convergence is not uniform... (I hope that it does not converge uniformly)
Other difficulty is that how do i show that this convergence is not uniform on any subset $B\subset [0,1]$ of measure $1$..
Help me to settle this..

Let $B \subseteq [0,1]$ any set of measure $1$, and $g_n := f_n \cdot \chi_B$. We show that $\|g_n\|_\infty \not\to 0$, which proves that $f_n$ does not converge to $\chi_{\{0\}}$ uniformly on $B$ (which is the only possible limit). To this end, note that as $\mu(B) + \mu([0,\frac 1n]) = 1+\frac 1n > \mu([0,1])$, we have $\mu(B \cap [0,\frac 1n]) > 0$. Hence $g_n = \chi_{B \cap [0,\frac 1n]}$ has value 1 on a set of positive measure, prooving $\|g_n\|_\infty = 1$. As $\|\chi_{\{0\}}\| = 0$, we cannot have $g_n \to \chi_{B \cap \{0\}}$ in $L^\infty$ (an hence not uniformly).
