$X$ a numerable set, $f$ a positive function, then $\mu(A) = \sum_{x\in A} f(x)$ $\sigma$-finite? Let $(X$, ${\mathscr P}(X), \mu)$ a measurable space, where $X$ is a numerable set, $f: X \rightarrow [0, \infty]$ a positive function. If we define the measure as $$\mu(A) = \sum_{x\in A} f(x), \text{for }A\subset X $$ is $\mu$ $\sigma$-finite? I feel like this shouldn't hold for every single positive function, any other than $\text{ }f(x) = \infty \text{ }\forall\text{ } x \in X$ which is pretty ridiculous, but I can't think of any other.

Actually, $\mu$ will be $\sigma$-finite if and only if $f(x)<+\infty$ for all $x\in X$.

Indeed, if $f(x)<+\infty$ for all $x\in X$, then $\\{\\{x\\},x\in X\\}$ is a countable partition of $X$ and each element as a finite measure.

If $\mu$ is $\sigma$-finite, there exists a sequence of subsets of $X$, say $\left(A_n\right)_{n\geqslant 1}$ such that $\mu\left(A_n\right)<+\infty$ and $\bigcup_{n\geqslant 1}A_n=X$. Let $x\in X$; there exists some $n$ such that $x\in A_n$. Then $$ f(x)=\mu\left(\\{x\\}\right)\leqslant \mu\left(A_n\right)<+\infty. $$