glitinosim3

2022-07-04

Given $\mathrm{\Omega }=\left(0,1\right)$, $\mathcal{A}$ is the Borel Sigma-Algebra restricted to $\mathrm{\Omega }$, and $\mu$ is the Lebesgue measure, let $f\left(\omega \right)=\frac{1}{\omega }$ and
$0<\omega <\frac{1}{n}$
$\frac{1}{n}\le \omega <1$
I have to prove that ${f}_{n}$ converges $\mu$-almost uniformly to $f$.
From what I learned in class, I have to show:

My idea was to let ${E}_{ϵ}=\left[ϵ-0.000000001,1\right)$. Then ${E}_{ϵ}^{c}=\left(0,ϵ-0.000000001\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mu \left({E}_{ϵ}^{c}\right)<ϵ.$
To prove uniform convergence, I need:
$\mathrm{\forall }\delta >0,\mathrm{\exists }{N}_{\delta }\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall }n>{N}_{\delta },\mathrm{\forall }\omega \in {E}_{ϵ},|{f}_{n}\left(\omega \right)-f\left(\omega \right)|<\delta .$
Since ${E}_{ϵ}=\left[ϵ-0.000000001,1\right),$ $\omega \ge ϵ-0.000000001$, so if $ϵ-0.000000001\ge \frac{1}{n}$, we get $|{f}_{n}\left(\omega \right)-f\left(\omega \right)|=0<\delta .$
This holds for all $n\ge \frac{1}{ϵ-0.000000001}$, so we let ${N}_{\delta }=\frac{1}{ϵ-0.000000001}$, proving the almost uniform convergence.
Is this argument correct?

Hayley Mccarthy

Expert

You can use the configuration of uniform convergence with the sup distance. Consider the partition ${E}_{m}^{c}=\left(0,1/m\right),{E}_{m}=\left[1/m,1\right)$ for $m\in \mathbb{N}$. We have that on ${E}_{m}^{c}$ the convergence is not uniform, but on ${E}_{m}$ it is as we have $\underset{\omega \in {E}_{m}}{sup}|{f}_{n}\left(\omega \right)-f\left(\omega \right)|=\left(m-1\right){\mathbf{1}}_{\left\{n:n. Notice this is valid for all $m\in \mathbb{N}$. Since $\mu \left(\left(0,1/m\right)\right)=1/m$, for arbitrary $\epsilon >0$ we can choose $m>1/\epsilon$ and the condition is satisfied.