In fact these conditions (measurable and integrable) are already sufficient. Indeed, let $A$ measurable. We have for a fixed $n$, denoting $E_n:=\left\{x,f(x)\leq n\right\}$: \begin{align} \int_A f(x)d\mu(x)&=\int_{A\cap E_n} f(x)d\mu(x)+\int_{A\cap E_n^c} f(x)d\mu(x)\\ &\leq n\mu(A)+\int_{E_n^c} f(x)d\mu(x)\\ &\leq n\mu(A)+\sum_{k=n}^{+\infty}(k+1)\mu(k\leq f < k+1),
\end{align} and since the series $\sum_{k=1}^{+\infty}k\mu(k\leq f is convergent, so is the series $\sum_{k=1}^{+\infty}(k+1)\mu(k\leq f, hence we can, given $\varepsilon>0$, find a $n$ such that $\sum_{k=n}^{+\infty}(k+1)\mu(k\leq f. Then for each $A$ measurable such that $\mu(A)\leq \frac{\varepsilon}{2n}$, we have $\int_A f(x)d\mu(x)\leq \varepsilon$.