$f_n$ $\in$ $L_2(\mu)$, the limit $ f \in L_2(\mu)$

Question

If $f_n \in L_2(\mu)$, $f_n\rightarrow f$ almost everywhere, this is not enough to conclude $f\in L_1(\mu)$.

But is it enough to conclude whether $f\in L_2(\mu)$ or $$\lim_{n \to \infty}\int_{R}{|f_n(x)-f(x)|}^2<\infty$$

What about the assumption change to $\sup\int_{R}{|f_n(x)|}^2d(\mu)<\infty$

Answer 1

No, it is not true. $$f_n(x) = \begin{cases} n & x \in [0,1/n)\\ 0 & \text{else}\end{cases}$$

Answer 2

• Take $\Bbb R$ with Lebesgue measur, and $f_n=\chi_{(0,n)}$. Then $f_n\in L^2$ and $f_n\to \chi_{(0,+\infty)}$, which is not (square)-integrable.

• The fact that $\sup_n\int_X f_n^2d\mu$ implies by Fatou's lemma that $f\in L^2$.

• But even in this case, we don't have necessarily convergence in $L^2$. Taking $f_n:=\sqrt n \chi_{(n,n+n^{—1})}$, we can see that $\int_Xf_n^2=1$ for all $n$ and $f_n\to 0$ almost everywhere.