Limit of integral over decreasing sequence of sets Let $(X, \mathcal{A}, \mu)$ be a measure space, $F \in L^1$, and $S_N$ sets such that $\mu(S_N) \to 0$. Can we conclude that $\int_{S_N} |F| d\mu \to 0$? Of course if the $S_N$ are contained it can be gotten from the monotone convergence property, but what if they are not?
 A: You can use that the measure $\nu$ defined by
$$
\nu(A)=\int_{A}|F| d\mu
$$
Is absolutely continuous with respect to $\mu$ to conclude wath you want.
Note that if for a measurable set $A$ we have $\mu(A)=0$ then the function $F\chi_A$ is 0 $\mu$-a.e. and then, we would have that $\nu(A)=0$ (by properties of the integral). Therefore, whenever we have $\mu(A)=0$, we will have $\nu(A)=0$. Another formulation of this is that given $\varepsilon>0$ we can find $\delta>0$ such that $\mu(A)<\delta$ implies $\nu(A)<\varepsilon $. To see this, suppose it is false and that there's a $\varepsilon>0$ such that for every $\delta>0$, there's a measurable set $A$  with $\nu(A)>\varepsilon$ even though $\mu(A)<\delta$. Construct a sequence of measurable sets $F_n$ such that $\mu(F_n)<2^{-n}$ and $\nu(F_n)>\varepsilon$ and call $E_n=\bigcup_{m=n}^\infty F_m$. Note that $E_{n+1}\subset E_n$ and that $\mu(E_n)\leq 2^{n-1}$. Hence, by properties of the measure
$$
\mu(\bigcap_{n=1}^\infty E_n)=\lim\limits_{n\to \infty}\mu(E_n)=0
$$
But we have that 
$$
\nu(\bigcap_{n=1}^\infty E_n)=\lim\limits_{n\to \infty}\nu(E_n)\geq \varepsilon
$$
And this is a contradiction to what we have shown. 
Now, using this, given $\varepsilon>0$ there exists a $\delta>0$ such that $\nu(A)\varepsilon$ if $\mu(A)<\delta$ and since $\mu(S_n)\to 0$ as $n\to \infty$ we have that there exists an $N$ such that $N\leq n$ implies $\mu(S_n)\leq \delta$ and hence, that $\nu(S_n)<\varepsilon$.
A: There's a simpler solution to the problem than the one above. Assume that $f$ is positive. Then what we want to show is that for every $\varepsilon > 0$ there is $\delta > 0$ with $\mu(S) < \delta \Rightarrow \int_S Fd \mu < \varepsilon$. To that end, let $f_n = f \wedge n$, and noting that $f_n$ converges monotonely to $f$, select $N$ with $\int f - f_N d \mu < \varepsilon/2$. Let $\delta = \varepsilon/2N$. Then if $\mu(S) < \delta$, we have
$$
\int_S f d \mu = \int_S f - f_N d \mu + \int_S f_N d \mu \leq \int f - f_N d \mu + N\int_S d \mu < \varepsilon
$$
