Use of Fatou's lemma to prove almost everywhere convergence I was trying to solve point 4. of this question based on an excercise taken from Probability Theory by A. Klenke (3rd version). Here is the link for completeness, but you can continue without reading as I am reporting here the main features below: Proof of Cramér-Lundberg inequality
I am not sure about my use of Fatou's lemma to show that $\lim_{n\to\infty} f_n=0$ a.e.
Let $f_n$ be a positive function (like $Z^\theta_n$ of the exercise in the link). We know that $\lim_{n\to\infty} E[\sqrt{f_n}]=0$ (also in the exercise). Moreover, $f_n$ is a martingale (as suggested by @jakobdt)
Using the Fatou's lemma and the hypothesis about the expected value in the last equality I get:
$0\leq\int \liminf_{n\to\infty} \sqrt{f_n} d\mu\leq\liminf_{n\to\infty}\int \sqrt{f_n}d\mu=0$.
So from $\int \liminf_{n\to\infty} \sqrt{f_n} d\mu = 0$ follows that $\liminf_{n\to\infty}\sqrt{f_n}=0 $ and that $\liminf_{n\to\infty}f_n=0$ a.e.
Up to this point, is it right? How can I continue to show $\lim_{n\to\infty} f_n=0$ a.e.?
Thank you.
 A: The statement is not true in general. For a counterexample, consider the sequence of functions $f_n : [0, 1]\to \mathbb{R}$ defined by
$$f_1=\boldsymbol{1}_{[0, 1]}, \quad f_2=\boldsymbol{1}_{[0,1/2]},\quad f_3=\boldsymbol{1}_{[1/2, 1]}, \quad f_4 =\boldsymbol{1}_{[0, 1/3]}, \quad f_5 =\boldsymbol{1}_{[1/3, 2/3]}, \quad f_6=\boldsymbol{1}_{[2/3, 1]}$$
and so on. See that $$\lim_{n\to\infty} \mathbb{E}(f_n)=\lim_{n\to\infty} \mathbb{E}\left(\sqrt{f_n}\right)=0$$
but $\forall x\in[0,1]$, $f_n(x)$ does not converge (so it actually converges nowhere, not just not almost everywhere).
Edit: I will do a quick edit to help OP finish the argument.
In the case that $\{f_n\}_{n=1}^\infty$ is a martingale (MG) then $\mathbb{E}\left(\sqrt{f_n}\right)\to0$ does imply that $f_n\overset{\text{a.s.}}{\to}0$ as $n\to\infty$. For $n\in\mathbb{N}$, $$\mathbb{E}\left(\sqrt{f_{n+1}} \middle|f_n,\dots,f_1\right)\le \sqrt{\mathbb{E}\left(f_{n+1} \middle|f_n,\dots,f_1\right)}=\sqrt{f_n}$$
where we used Jensen's inequality for concave functions and the MG property. Thus $\{\sqrt{f_n}\}_{n=1}^\infty$ is a super-martingale (superMG). Since $\mathbb{E}\left(\sqrt{f_n}\right)\ge 0$, it follows that $$\sup_{n\in\mathbb{N}}\mathbb{E}\left|\sqrt{f_n}\right|=\sup_{n\in\mathbb{N}}\mathbb{E}\left(\sqrt{f_n}\right)=\mathbb{E}\left(\sqrt{f_1}\right)<\infty$$
Using the superMG convergence theorem, there exists an integrable function $f$ such that $\sqrt{f_n}\overset{\text{a.s.}}{\to}f$ as $n\to\infty$. Using Fatou's lemma and the argument made by OP,
$$f=\liminf_{n\to\infty}\sqrt{f_n}\overset{\text{a.s.}}{=}0$$
and thus $f=0$ almost surely. Therefore, $f_n\to0$ almost surely as $n\to\infty$ as well.
A: As $(f_n)$ is a martingale we know that $(\sqrt{f_n})$ is a supermartingale (concave transformation of a martingale). Since $\sup_{n}E[(\sqrt{f_n})^-]=0<\infty$ we know that $\sqrt{f_n}$ converges almost surely to some limit $g$. You already know that $E[\sqrt{f_n}]\to0$, i.e. that $\sqrt{f_n}\to0$ in $L^1$. Hence, $g=0$ almost surely. We conclude that $f_n=(\sqrt{f_n})^2\to0$ almost surely.
