Does convergence of integrals on closed intervals imply convergence of integrals on measurable sets? I have a series of functions $(f_n)$, and I know that for any closed interval $[a,b]$ in R $\lim_{n \to \infty} \int_a^b f_n = 0$.
Is it true that for any lebesgue-measurable set $A$ in [0, 1]
$\lim_{n \to \infty} \int_A f_n = 0$?
I think it is possible to show it using the regularity of lebesgue measure but I'm not sure how.
(I talk about lebesgue integrals.)
 A: No. Let $c_n$ be a strictly increasing sequence in $[0,1]$ which converges, say, to 1, (e.g. $c_n = 1-1/n$) and consider
$$f_n(x) = \begin{cases} -(c_{n+1}-c_n)^{-1} & c_{n} < x < c_{n+1} \\
(c_{n+2}-c_{n+1})^{-1}, & c_{n+1} < x < c_{n+2} \\
0, & \text{otherwise}.\end{cases}$$
Let $0 \le a < b \le 1$.  If $b < 1$ then for all large enough $n$, we have $c_n > b$ and therefore $f_n \equiv 0$ on $[a,b]$, so $\int_a^b f_n \to 0$.  If $b = 1$ then for all large enough $n$ we have $c_n > a$ and we can again compute that $\int_a^b f_n = 0$, so again $\int_a^b f_n \to 0$.
But take $A = [c_1, c_2] \cup [c_3, c_4] \cup [c_5, c_6] \cup \dots$.  You can easily see that $\int_A f_n = (-1)^n$ which does not converge to 0.
It would be true under the additional assumption that 
$$\sup_n \int |f_n|^p < \infty$$
and in particular if the $f_n$ were uniformly bounded.  Note that we have $\int f_n g \to 0$ for any "step function" $g = \sum_{i=1}^m d_i 1_{[a_i, b_i]}$.  Step functions are dense in $L^q([0,1])$ where $\frac{1}{p} + \frac{1}{q} = 1$ (this uses the regularity of Lebesgue measure) and so for any $A$ and any $\epsilon$, there is a step function $g$ with $\|1_A - g\|_{L^q} < \epsilon$.  Now let $C = \sup_n \int |f_n|^p$ and write
$$\begin{align*}
\left|\int_A f_n\right| &\le \left|\int f_n \cdot (1_A - g)\right| + \left|\int f_n g \right| \\ &\le \|f_n \|_{L^p} \|1_A - g\|_{L^q} + \left|\int f_n g \right| && \text{Holder's inequality} \\
&\le C \epsilon + \left|\int f_ng \right|.\end{align*}$$
Since $\int f_ng \to 0$, we conclude $\limsup_{n \to \infty} \int_A f_n \le C\epsilon$, and $\epsilon$ was arbitrary.
This breaks down when $p=1$ because step functions are not dense in $L^\infty$.
