In $(\mathbb{Z}, \mathcal{P}(\mathbb{Z}), \mu)$, convergence in measure is equivalent to uniform convergence. 
In $(\mathbb{Z}, \mathcal{P}(\mathbb{Z}), \mu)$,, where $\mu(A)=\operatorname{card}(A)$ for $A \subset \mathbb{Z}$, show that if $f, f_1, f_2,\dots : \mathbb{Z} \rightarrow \mathbb{R}$ then $(f_n)\rightarrow f$ in measure if and only if $(f_n)\rightarrow f$ uniformly.

The definition of convergence in measure is that, given $\varepsilon >0$,
$$\lim_n \mu (\{x \in X: |f_n(x)-f(x)| \geq \varepsilon \}) = 0$$
Now I need to go from that to show that there is some $N \in \mathbb{N}$ such that 
$$|f_n(x)-f(x)|<\varepsilon$$
whenever $n \geq N$, using the fact that $\mu(A) = \operatorname{card}(A)$. Can someone give my a hint on how to proceed? Thank you in advance.
 A: Hint: if $\lim a_n = 0$ then there exists $N$ such that for all $n\ge N$, $|a_n| < 1/2$. Also if $\text{card} (A) < 1/2$, then $A = \varnothing$.
A: Hint:
The cardinality of set $\{k\in\mathbb N:|f_n(k)-f(k)|\geq\epsilon\}$ takes values in $\{0,1,\dots,\infty\}$, so can only tend to $0$ if for $n$ large enough it takes value $0$.
A: The main fact at work here is that your measure space does not have sets of arbitrarily small positive measure and the measure of every nonempty set is positive. Given a general measure space $(X,\mathcal{M},\mu)$ with this property, denote by
$$\alpha:=\inf\{\mu(A) \mid A\in\mathcal{M},\,A\ne\emptyset\}.$$
Then our assumption gives that $\alpha>0$.
Suppose $f_n\to f$ in measure and $\varepsilon>0$. Since $\mu[|f_n-f|>\varepsilon]\to0$, there exists $N\in\mathbb{N}$ such that
$$
\mu[f_n-f|>\varepsilon] < \alpha/2
$$
for every $n\geq N$. This implies that $[|f_n-f|>\varepsilon]=\emptyset$ for every $n\geq N$, which is just another way of saying that $\|f_n-f\|_\infty \leq \varepsilon $ for all $n\geq N$. Therefore $f_n\to f$ uniformly.
The converse holds in general, for if $f_n\to f$ uniformly, then for every $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that $[|f_n-f|>\varepsilon]=\emptyset$ whenever $n\geq N$.
