Lucia Grimes

2022-07-02

In measure theory I encountered Egorov's theorem which states that if $\left(X,\mathcal{S},\mu \right)$ is a measure space such that $\mu \left(X\right)<\mathrm{\infty }$ i.e. $\mu$ is a finite measure.If $\left({f}_{n}\right)$ 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 $ϵ>0$,there exists ${E}_{ϵ}\subset X$ such that $\mu \left({E}_{ϵ}\right)<ϵ$ and ${f}_{n}\to f$ uniformly on $X-{E}_{ϵ}$.
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 $\mathrm{\exists }E\subset X$ such that $\mu \left(E\right)=0$ 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".

Sanaa Hinton

Expert

Here is a counterexample: ${u}_{n}\left(x\right)={x}^{n}$ 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 $\left[0,1\right]\mathrm{\setminus }S$: 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 ${S}_{m}=\left(1-1/m,1\right]$ such that $\mu \left({S}_{m}\right)\to 0$, and ${u}_{n}$ converges uniformly on $\left[0,1\right]\mathrm{\setminus }{S}_{m}$ for all $m$.