In measure theory I encountered Egorov's theorem which states that if $(X,\mathcal S,\mu)$ is a measure space such that $\mu(X)<\infty$ i.e. $\mu$ is a finite measure.If $(f_n)$ be a sequence of measurable functions on $X$ converging pointwise to $f:X\to \mathbb R$,then $f_n$ is almost uniformly convergent to $f$.
Now in the book the definition of almost uniform convergence is the following:
$f_n\to f$ almost uniformly if for each $\epsilon>0$,there exists $E_\epsilon\subset X$ such that $\mu(E_\epsilon)<\epsilon$ and $f_n\to f$ uniformly on $X-E_\epsilon$.
Now in some books I have seen that almost everywhere means outside a measure $0$ set.
So the definition of almost uniformly convergent should have been $f_n\to f$ almost uniformly if $\exists E\subset X$ such that $\mu(E)=0$ and $f_n\to f$ uniformly on $X-E$.
But unfortunately the definition is not so.In fact the latter condition is stronger.I want to know why the former is taken as a definition and what the problem with the latter one is.I want to understand where I am making mistake in understanding the word "almost".
Best Answer
Your alternate definition wouldn't work.
Here is a counterexample: $u_n(x)=x^n$ on $[0,1]$, which converges pointwise to $0$ if $x<1$ and $1$ otherwise. This sequence doesn't converge uniformly on $[0,1]$, nor on the half-open set $[0,1)$, and you can't find a set $S$ of measure $0$ such that convergence is uniform on $[0,1]\backslash S$: you would still need values of $x$ arbitrarily close to $1$.
However, $u_n$ converges uniformly on every compact interval that doesn't contain $1$, so you can find $S_m=(1-1/m,1]$ such that $\mu(S_m)\to0$, and $u_n$ converges uniformly on $[0,1]\backslash S_m$ for all $m$.