Sequence of sets converge to zero in measure if integral converges to zero. 
Let $f:[0,1]\to(0,\infty)$ be an integrable function and let
$\{F_n\}_n\subseteq [0,1]$ be a sequence of measurable sets such that
\begin{equation*}
     \int_{F_n}f(x)dx\to 0 \text{ as } n\to\infty. \end{equation*}
Show that $m(F_n)\to 0$ as $n\to\infty$.

(ATTEMPT) Notice that
\begin{equation*}
    \int_0^1 f(x)dx=\int_{F_n}f(x)dx+\int_{F_n^c}f(x)dx, \quad \forall n\in\mathbb{N}
\end{equation*}
and therefore,
\begin{align*}
    \int_0^1 f(x)dx&=\lim_{n\to\infty} \int_0^1 f(x)dx=\lim_{n\to\infty}\int_{F_n}f(x)dx+\lim_{n\to\infty}\int_{F_n^c}f(x)dx\\
    &=\lim_{n\to\infty}\int_{[0,1]-F_n}f(x)dx \text{ by assumption}.
\end{align*}
By the Dominated Convergence Theorem, since f is integrable, we have that
\begin{equation*}
    \lim_{n\to\infty}\int_{[0,1]-F_n}f(x)dx=\int_{0}^1 \lim_{n\to\infty}f(x)\chi_{[0,1]-F_n}(x)dx. 
\end{equation*}
Hence,
\begin{align*}
    \int_0^1 \left[f(x)-\lim_{n\to\infty}f(x)\chi_{[0,1]-F_n}(x)\right]dx=0.
\end{align*}
Since the function under the integral is non-negative, we must have that
\begin{align*}
    f(x)=\lim_{n\to\infty}f(x)\chi_{[0,1]-F_n}(x) \text{ or } \lim_{n\to\infty}\chi_{[0,1]-F_n}(x)=1.
\end{align*}
Then, we must have that $m(F_n)\to 0$ as $n\to\infty$.
I am unsure about the last step, and I will also appreciate any comments on any part of the proof.
 A: You can't apply the dominated convergence theorem since you don't know whether the pointwise limit $\lim_{n \to \infty}f(x)\chi_{[0, 1] - F_n}(x)$ exists for all (or a.e.) $x$.
Edit: Full Solution. First assume that $f = \sum_{k = 1}^{\infty}c_k\chi_{A_k}$, where each $c_k > 0$, $\{A_k\}$ are disjoint and measurable, and $\bigcup_{k = 1}^{\infty}A_k = [0, 1]$. The hypothesis says that
\begin{align}
\int_{F_n}f\,dx &= \int_{[0, 1]}\chi_{F_n}f\,dx \\
&= \int_{[0, 1]}\sum_{k = 1}^{\infty}c_k\chi_{F_n \cap A_k}\,dx \\
&= \sum_{k = 1}^{\infty}c_km(F_n \cap A_k) \to 0 \text{ as } n \to \infty.
\end{align}
Thus for each $k \in \mathbb{N}$, $m(F_n \cap A_k) \to 0$ as $n \to \infty$. We would like to show that
\begin{equation}
\lim_{n \to \infty}m(F_n) = \lim_{n \to \infty}\sum_{k = 1}^{\infty}m(F_n \cap A_k) = 0.
\end{equation}
We have $\lim_{n \to \infty}\sum_{k = 1}^{\infty}m(F_n \cap A_k) = \lim_{n \to \infty}\int_{\mathbb{N}}m(F_n \cap A_k)d\mu(k)$, where $\mu$ denotes the counting measure on $\mathbb{N}$. For all $n, k \in \mathbb{N}$ we have $|m(F_n \cap A_k)| \leq m(A_k)$, and $\int_{\mathbb{N}}m(A_k)\,d\mu(k) = \sum_{k = 1}^{\infty}m(A_k) = m([0, 1]) = 1 < \infty$. Thus by DCT, $\lim_{n \to \infty}\sum_{k = 1}^{\infty}m(F_n \cap A_k) = \sum_{k = 1}^{\infty}\lim_{n \to \infty}m(F_n \cap A_k) = 0$. Thus the theorem holds in this special case.
Now suppose $f$ satisfies the original hypotheses of the theorem. Define $\phi \colon [0, 1] \to (0, \infty)$ by
$$\phi = \sum_{k = 0}^{\infty}\frac{1}{2(k + 1)}\chi_{\{\frac{1}{k} > f \geq \frac{1}{k + 1}\}}.$$
$0 < \phi < f$, so $\phi$ satisfies the hypotheses of the theorem. Thus we can apply the special case to conclude that $m(F_n) \to 0$ as $n \to \infty$.
A: Here is a direct solution that does not require approximation by simple functions. I think it makes things a little more transparent.
Let $A_k=\{x\in [0,1]\mid f(x)\geq 1/k\}$. Since $f>0$, we have $[0,1]=\bigcup_k A_k$. Thus for every $\epsilon>0$ there exists $k\in\mathbb N$ such that $m([0,1]\setminus A_k)<\epsilon$.
We have
$$
m(F_n)=m(F_n\setminus A_k)+m(F_n\cap A_k)\leq \epsilon+k\int_{F_n\cap A_k}f\,dx\leq k\int_{F_n}f\,dx\to \epsilon.
$$
Hence $\limsup_n m(F_n)<\epsilon$. Since $\epsilon>0$ was arbitrary, we conclude $\lim_n m(F_n)=0$.
