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.