Conditions under which the Limit for “Measure $\to 0$” is $0$

Let $\mu$ be a probability measure on $X$, so that $\int_X \mu(dx) = 1$.

Say under which conditions on the function $f: X \rightarrow \mathbb{R}_{> 0} \$ (that is measurable and integrable) we have that

$$\lim_{\mu(A) \rightarrow 0 } \int_A f(x) \mu(dx) = 0$$

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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 $\sum_{k=1}^{+\infty}k\mu(k\leq f<k+1)$ is convergent, so is the series $\sum_{k=1}^{+\infty}(k+1)\mu(k\leq f<k+1)$, hence we can, given $\varepsilon>0$, find a $n$ such that $\sum_{k=n}^{+\infty}(k+1)\mu(k\leq f<k+1)\leq \frac{\varepsilon}2$. Then for each $A$ measurable such that $\mu(A)\leq \frac{\varepsilon}{2n}$, we have $\int_A f(x)d\mu(x)\leq \varepsilon$.
I write $E_n^c$ has the union $\bigcup_{k\geq n}\{x,k\leq f(x)<k+1\}$, and on this set $f(x)<k+1$. We have $\sum_{k=1}^{+\infty}k\mu(k\leq f<k+1)\leq \int f d\mu<\infty$ and the series $\sum_{k=0}^{+\infty}\mu(k\leq f<k+1)$ is convergent since it's the measure of the set $\{f\geq 0\}$, which is finite since the measure of $X$ is finite. –  Davide Giraudo Mar 10 '12 at 9:55
Another way to see this is to note that if $A_n$ is a sequence of measurable sets with $\mu(A_n) \to 0$, then $f 1_{A_n} \to 0$ in measure. Since $|f 1_{A_n}| \le |f|$ and $f$ is integrable, an appropriate version of the dominated convergence theorem shows that $\int 1_{A_n} f\,d\mu \to 0$.