Is the application of Lebesgue DCT valid here?

Question

Problem:

Let $f_n: [0,1] \to \mathbb{R}$ be a sequence of measurable functions.

Suppose that $\int_{0}^{1}|f_n(x)|^2 ~ dx \le 1$ for $n \in \mathbb{N}$ and $f_n$ converges to $0$ a.e.

Show that $\lim_{n \to \infty} \int_{0}^{1} f_n(x) ~ dx = 0$.

Question: Is the following solution correct? If not, how can it be fixed?

Proposed Solution:

We have $\|f_n\|_{L^2} \le 1$ $\forall n \in \mathbb{N}$ and the measure space is sigma-finite. So $\|f_n\|_{L^1} \le \|f_n\|_{L^2} \le 1$ and $f_n$ is bounded above by the integrable function $g(x) = 1$ for every $n$. By the pointwise-a.e. convergence, $f \equiv 0$, and $|f| \le g$. So by DCT, the result follows.

Answer 1

Here is a way to solve the problem. Since $f_n\to 0$ a.e, and the measure space is finite, you have $f_n\to 0$ in probability. Given $\epsilon>0$, let $A_n=\{|f_n|>\epsilon\}$. Then $$ \int_0^1 |f_n|\,dx=\int_{A_n} |f_n|\,dx + \int_{[0,1]\setminus A_n}|f_n|\,dx\le \int_0^1 f_n(x){\bf 1}_{A_n}(x)\,dx + \epsilon $$ Now apply Cauchy-Schwarz to that last integral.

Answer 2

The proof is false. DCT requires that the function be dominated in the sense that $|f_{n}| \leq g$.

Also, the inequalities that you use don't apply in a sigma finite measure space. They apply in a finite measure space! They are false over the entire real line!