Lucia Grimes

Answered

2022-07-02

In measure theory I encountered Egorov's theorem which states that if $(X,\mathcal{S},\mu )$ is a measure space such that $\mu (X)<\mathrm{\infty}$ 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 $\u03f5>0$,there exists ${E}_{\u03f5}\subset X$ such that $\mu ({E}_{\u03f5})<\u03f5$ and ${f}_{n}\to f$ uniformly on $X-{E}_{\u03f5}$.

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 $\mu (E)=0$ 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".

Now in the book the definition of almost uniform convergence is the following:

${f}_{n}\to f$ almost uniformly if for each $\u03f5>0$,there exists ${E}_{\u03f5}\subset X$ such that $\mu ({E}_{\u03f5})<\u03f5$ and ${f}_{n}\to f$ uniformly on $X-{E}_{\u03f5}$.

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 $\mu (E)=0$ 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".

Answer & Explanation

Sanaa Hinton

Expert

2022-07-03Added 15 answers

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

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

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