A question about sequences of integrable functions Say $f_n$ is some sequence of Lebesgue integrable maps $[0,1] \to \mathbb R$ such that for all $x\in [0,1]$:
$$ \lim_{n \to \infty} f_n (x) = 0$$
That is, the pointwise limit is the zero function. 
Now consider the sequences 
$$ \int_0^1 f_n(x)dx$$
and 
$$ \int_0^1 |f_n(x)|dx$$
I believe that if the pointwise limit function is zero then these should also tend to zero but I'm not 100% sure. 
Especially, I'm not sure how to argue it because they are only integrable and not necessarily continuous.

Does anyone have any insights into this?

 A: Nope, not true, even for everywhere pointwise convergence: Consider
$$
f_n(x) = \begin{cases}
n^2x & 0\leq x\leq 1/n\\
2n-n^2x & 1/n<x\leq 2/n\\
0 & x>2/n
\end{cases}.
$$
(Note that all $f_n$ are continuous…). Also look up Lebesgue's dominated convergence theorem.
A: An easy counterexample in the case of almost everywhere convergence, since I've already written it. Dirk took care of the question better.
I also recommend to read something like this Tao's post  about modes of convergence, to get a better idea of what's going on.

Take $f_n(x)=n\chi_{[0,\frac{1}{n}]}(x)$. Then $f_n \to f$ almost everywhere and 
$$
\int_0^1 |f_n(x)|\,dx=\int_0^1 f_n(x)\,dx=n\int_0^1 \chi_{[0,\frac{1}{n}]}(x)\, dx= n \mu\left(\left[0,\frac{1}{n}\right]\right)=1,
$$ 
for any $n$ and where we denoted by $\mu$ the Lebesgue measure. 
A: The answer is no. Here is a counterexample that $f_n(x)\to0$, but $\int_0^1 f_n(x)dx\to\infty$ as $n\to\infty$.
Let $f_n(x)=n^2xe^{-nx^2}$ on $[0,1]$. Then for any $x\in[0,1]$
$$
\lim_{n \to \infty} f_n (x) = 0
$$
 But
$$
\int_0^1 f_n(x)dx=\frac{n}{2}(1-e^{-n})\to\infty
$$
A: Or look at $f_n(x) = n^2x^n(1-x).$
