True/false : If $f_n(x)\to 0$ almost everywhere , then $\int_{a}^{b}f_n(x)dx \to 0$

Question

Is the following sttaement is true/false ?

let {$f_n$} be sequence of integrable functions defined on an interval $[a,b]$.

If $f_n(x)\to 0$ almost everywhere , then $\int_{a}^{b}f_n(x)dx \to 0$

My attempt : i know that $m(\mathbb{Q}) = 0$ and $f(\mathbb{Q}) =0$ is almost everywhere because the $f$ is not going to $0$ and measure is $0$ in $[0,1]$

Now consider the function $f_n(x)$, on the interval $[0,1]$. Let $\{r_1, r_2, \dots\}
$ is the enumerarion of rationals on $[0,1]$.

$$f_n(x) = \begin{cases} 1 & \text{x = $r_1, r_2, \dots r_n$} \\ 0 & \text{elsewhere} \end{cases}$$

Thus $f_n(x) \rightarrow f(x)$, which is $0$ almost everywhere

$$f(x) = \begin{cases}1 & x \in \mathbb{Q}\cap [0,1] \\0 & x \in [0,1] – \mathbb{Q}\cap [0,1]\end{cases}$$

But $\int_0^1 f(x)$ does not exists.

so the given statement is false

Is its true ?

Answer 1

The usual counterexample is something like $$f_n(x) = \begin{cases}n & x \in (0, 1/n] \\ 0 & \text{otherwise}\end{cases}$$