If $f\in M^{+}(X,$ X $)$ and $\int f\, d\mu<{\infty}$, then the set $N = \{x\in X: f(x)>0\}$ is $\sigma$-finite (that is, there exists a sequence $(F_n)$ in X such that $N \subset \bigcup F_{n}$ and $\mu(F_n)<\infty$).
Consider the sets $F_n = \{x \in X: f(x) > \frac{1}{n}\}$. Can you see why $F_n$ must be a set of finite measure for $\int f d\mu$ to be finite?