Let $\{f_n\}_{n=1}^{\infty}$ and $f$ be integrable functions on $[0, 1]$ such that $$\lim_{n \to \infty}\int_0^1|f_n(x) - f(x)| dx = 0$$
Is there exists an example which shows that following statement doesn't hold good?
$f_n(x) \to f(x)$, as $n \to \infty$, for almost every $x \in [0, 1]$.
Consider the indicator function
$$1_{(a,b)}(x) = \begin{cases}1, \,\, x \in(a,b)\\0, \,\,x \notin (a,b) \end{cases} $$
Take $f(x) = 0$ for all $x$ and
$$f_1 = 1_{(0,1/2)}\, , \,\,f_2 = 1_{(1/2,1)} \,, \,\, f_3 = 1_{(0,1/4)} \, , \,\, f_4 = 1_{(1/4,1/2)}\, , \, \, f_5 = 1_{(1/2,3/4)} \, , f_6 = 1_{(3/4,1)} \, , \ldots $$
Now show $\int_0^1|f_n(x)| \, dx \to 0$, but for almost every $x \in (0,1)$ and for every $n \in {N}$ there exists $n_1, n_2 > n$ such that $f_{n_1}(x) = 1$ and $f_{n_2}(x) = 0$.