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
$$ \lim_{\mu(A) \rightarrow 0 } \int_A f(x) \mu(dx) = 0 $$
 A: 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 $\int 1_{A_n} f\,d\mu \to 0$.
A: 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 $\int_A f(x)d\mu(x)\leq \varepsilon$.
