1
$\begingroup$

Problem 2.4.42 - Let $\mu$ be counting measure on $\mathbb{N}$. Then $f_n\rightarrow f$ in measure if and only if $f_n\rightarrow f$ uniformly.

Attempted proof - Suppose that $\mu$ is a counting measure on $\mathbb{N}$ and $f_n\rightarrow f$ in measure for all $n\in\mathbb{N}$. Given $\epsilon\in (0,\infty)$ there exists an $N\in\mathbb{N}$ such that $$\mu\left(\{x\in\mathbb{N}:|f_n(x) - f(x)| \geq \epsilon\}\right) < 1$$ for all $n\in\mathbb{N}$ with $n\geq N$. This implies that $\{x\in\mathbb{N}:|f_n(x) - f(x)\geq \epsilon\}\ = \emptyset$. Hence we have for all $\epsilon\in (0,\infty)$ there exists an $N\in\mathbb{N}$ for every $x\in X$, $|f_n(x) - f(x)| < \epsilon$ for all $n\geq N$. Thus, $f_n\rightarrow f$ uniformly.

Conversely, suppose $f_n\rightarrow f$ uniformly. If $\epsilon\in (0,\infty)$ there exists an $N\in\mathbb{N}$ for every $x\in X$, $|f_n(x) - f(x)| < \epsilon$ for all $n\geq N$. In particular $$\mu\left(\{x:|f_n(x)-f(x)|\geq\epsilon\}\right) = 0$$ for all $n\in\mathbb{N}$ with $n\geq N$, thus $f_n\rightarrow f$ in measure.

I am pretty sure this is correct, although a tad messy I may have made some mistakes. Any suggestions is greatly appreciated.

$\endgroup$
  • 1
    $\begingroup$ I am voting to close as off-topic because MSE is not a suitable place for posting solutions to every problem / proofs for every theorem in a textbook. $\endgroup$ – user296602 Jul 2 '16 at 4:21
  • 1
    $\begingroup$ @T.Bongers I fail to understand how my problems/theorems that I post to help improve my proof writing skills and assist in providing detailed solutions to others within the MSE community can bother you. You can simply choose not to read my questions I post instead of commenting negatively to people that are trying to use MSE as a way to improve ones skills and potentially help others. $\endgroup$ – Wolfy Jul 2 '16 at 4:29
  • 1
    $\begingroup$ If your questions are meant to be about your own development in proof-writing, then the sheer volume of them means that you should not be seeking help from a website, but rather a professor or mentor; you've posted literally dozens of these. If your questions are meant to be about compiling a solutions manual to Folland, then start your own blog to do so. $\endgroup$ – user296602 Jul 2 '16 at 4:36
  • 1
    $\begingroup$ Those are not my intentions I am trying to help myself and others which MSE is for. Kindly disregard my posts if they do not help you. $\endgroup$ – Wolfy Jul 2 '16 at 5:02
1
$\begingroup$

Contratulations! Your proof is correct. I have just copied it here to minor improvement in wording and making the final step clearer.

Problem 2.4.42 - Let $\mu$ be counting measure on $\mathbb{N}$. Then $f_n\rightarrow f$ in measure if and only if $f_n\rightarrow f$ uniformly.

Proof - Suppose that $\mu$ is a counting measure on $\mathbb{N}$ and $f_n\rightarrow f$ in measure. Given $\epsilon\in (0,\infty)$ there exists an $N\in\mathbb{N}$ such that $$\mu\left(\{x\in\mathbb{N}:|f_n(x) - f(x)| \geq \epsilon\}\right) < 1$$ for all $n\in\mathbb{N}$ with $n\geq N$.

This implies that $\{x\in\mathbb{N}:|f_n(x) - f(x)\geq \epsilon\}\ = \emptyset$. Hence we have for all $\epsilon\in (0,\infty)$ there exists an $N\in\mathbb{N}$ such that, for every $x\in X$, $|f_n(x) - f(x)| < \epsilon$ for all $n\geq N$. Thus, $f_n\rightarrow f$ uniformly.

Conversely, suppose $f_n\rightarrow f$ uniformly. If $\epsilon\in (0,\infty)$ there exists an $N\in\mathbb{N}$ such that, for every $x\in X$, $|f_n(x) - f(x)| < \epsilon$ for all $n\geq N$. In particular $$\mu\left(\{x:|f_n(x)-f(x)|\geq\epsilon\}\right) = 0$$ for all $n\in\mathbb{N}$ with $n\geq N$.

So, given $\epsilon\in (0,\infty)$ and $\delta >0$ there exists an $N\in\mathbb{N}$ such that $$\mu\left(\{x\in\mathbb{N}:|f_n(x) - f(x)| \geq \epsilon\}\right)=0 < \delta$$ for all $n\in\mathbb{N}$ with $n\geq N$.

Thus $f_n\rightarrow f$ in measure.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.