glitinosim3

Answered

2022-07-04

Given $\mathrm{\Omega}=(0,1)$, $\mathcal{A}$ is the Borel Sigma-Algebra restricted to $\mathrm{\Omega}$, and $\mu $ is the Lebesgue measure, let $f(\omega )=\frac{1}{\omega}$ 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:

$\mathrm{\forall}\u03f5>0,\mathrm{\exists}{E}_{\u03f5}\in \mathcal{A},\mu ({E}_{\u03f5}^{c})<\u03f5\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{f}_{n}\text{converges uniformly to}f\text{on}{E}_{\u03f5}.$

My idea was to let ${E}_{\u03f5}=[\u03f5-0.000000001,1)$. Then ${E}_{\u03f5}^{c}=(0,\u03f5-0.000000001)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mu ({E}_{\u03f5}^{c})<\u03f5.$

To prove uniform convergence, I need:

$\mathrm{\forall}\delta >0,\mathrm{\exists}{N}_{\delta}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}n>{N}_{\delta},\mathrm{\forall}\omega \in {E}_{\u03f5},|{f}_{n}(\omega )-f(\omega )|<\delta .$

Since ${E}_{\u03f5}=[\u03f5-0.000000001,1),$ $\omega \ge \u03f5-0.000000001$, so if $\u03f5-0.000000001\ge \frac{1}{n}$, we get $|{f}_{n}(\omega )-f(\omega )|=0<\delta .$

This holds for all $n\ge \frac{1}{\u03f5-0.000000001}$, so we let ${N}_{\delta}=\frac{1}{\u03f5-0.000000001}$, proving the almost uniform convergence.

Is this argument correct?

$0<\omega <\frac{1}{n}$

$\frac{1}{n}\le \omega <1$

I have to prove that ${f}_{n}$ converges $\mu $-almost uniformly to $f$.

From what I learned in class, I have to show:

$\mathrm{\forall}\u03f5>0,\mathrm{\exists}{E}_{\u03f5}\in \mathcal{A},\mu ({E}_{\u03f5}^{c})<\u03f5\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{f}_{n}\text{converges uniformly to}f\text{on}{E}_{\u03f5}.$

My idea was to let ${E}_{\u03f5}=[\u03f5-0.000000001,1)$. Then ${E}_{\u03f5}^{c}=(0,\u03f5-0.000000001)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mu ({E}_{\u03f5}^{c})<\u03f5.$

To prove uniform convergence, I need:

$\mathrm{\forall}\delta >0,\mathrm{\exists}{N}_{\delta}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}n>{N}_{\delta},\mathrm{\forall}\omega \in {E}_{\u03f5},|{f}_{n}(\omega )-f(\omega )|<\delta .$

Since ${E}_{\u03f5}=[\u03f5-0.000000001,1),$ $\omega \ge \u03f5-0.000000001$, so if $\u03f5-0.000000001\ge \frac{1}{n}$, we get $|{f}_{n}(\omega )-f(\omega )|=0<\delta .$

This holds for all $n\ge \frac{1}{\u03f5-0.000000001}$, so we let ${N}_{\delta}=\frac{1}{\u03f5-0.000000001}$, proving the almost uniform convergence.

Is this argument correct?

Answer & Explanation

Hayley Mccarthy

Expert

2022-07-05Added 19 answers

You can use the configuration of uniform convergence with the sup distance. Consider the partition ${E}_{m}^{c}=(0,1/m),{E}_{m}=[1/m,1)$ 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 $\underset{\omega \in {E}_{m}}{sup}|{f}_{n}(\omega )-f(\omega )|=(m-1){\mathbf{1}}_{\{n:n<m\}}(n)\stackrel{n\to \mathrm{\infty}}{\to}0$. Notice this is valid for all $m\in \mathbb{N}$. Since $\mu ((0,1/m))=1/m$, for arbitrary $\epsilon >0$ we can choose $m>1/\epsilon $ and the condition is satisfied.

Most Popular Questions