Let $(X,\mathcal{A}, \mu)$ be a measure space. Prove:
$\mu$ is $\sigma$-finite $\iff \exists$ function $f \in \mathcal{L}^{1}(\mu)$ with $f(x)>0$ for all $x \in X$
My ideas
$"\Leftarrow"$
Let $f \in \mathcal{L}^{1}(\mu)$ with $f(x)>0$, $\forall x \in X$. So, $f$ is measurable. This implies that for a $B_{n}$ defined as $B_{n}:=\{f>\frac{1}{n}\}$ which is measurable, so $\in \mathcal{A}$. By definition, $\{f>0\}=\bigcup_{n\in \mathbb N}B_{n}\in \mathcal{A}$, but since $f > 0, \forall x \in X$ then $X\subseteq\bigcup_{n\in \mathbb N}B_{n}$ and $\mu(B_{n})<\infty, \forall n \in \mathbb N$, since $\int_{X}fd\mu < \infty$$\Rightarrow \mu$ is $\sigma-$finite.
$"\Rightarrow"$ I have no idea how to define this function, particularly as $f>0$, $\forall x \in X$