# Doubts regarding almost uniform convergence

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".

• For continuous functions, say on $\mathbb R$, uniform convergence outside a set of measure $0$ implies uniform convergence everywhere. So the concept is not very useful. Mar 28, 2022 at 11:07

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 $$m$$.