Real Analysis, Folland Problem 2.4.42, counting measure with convergence in measure 
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.
 A: 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.
