If a martingale $M$ has $\sup_{t\geq 0} \mathbb{E}(M_t^2)<\infty$, then how do we prove that we also have $\mathbb{E}(\sup_{t\geq 0}M_t^2)<\infty$?
The hint in the book is to use Doob's $L^p$ inequality, to obtain $\mathbb{E}(\sup_{N\geq t\geq 0}M_t^2)<4\mathbb{E}(M_N^2)$, and then take the limits (w.r.t N) and use Fatou's lemma.
My doubt resides in the use of the limit and Fatou's lemma.
The (reverse) Fatou's lemma allows, under certain conditions, to state: $\limsup_{N\rightarrow \infty}\mathbb{E}(M_N^2)\leq \mathbb{E}(\limsup_{N\rightarrow \infty} M_N^2)$.
However, $\lim_{N\rightarrow \infty}\sup_{N\geq t\geq 0}M_N^2$ is different from $\limsup$, since it's $\lim_{t\rightarrow \infty}\sup_{N\geq t\geq 0}M_N^2 = \inf_{t\geq 0}\sup_{N\geq t} M_N^2$.
Also, it seems we would want the inequality in the opposite direction, i.e., $\mathbb{E}(\limsup_{N\rightarrow \infty} M_N^2)\leq \lim_{N\rightarrow \infty}\sup_{N\geq t \geq 0}\mathbb{E}(M_N^2) \leq \sup_{t\geq 0} \mathbb{E}(M_t^2)\leq \infty$.
I'm not really sure how to use Fatou's lemma in this context...
Let $f_{N}:=\sup_{N\geq t\geq 0}M_{t}^{2}$, then by (liminf)-Fatou's lemma we get
$$E[\liminf_{N}f_{N}]\leq \liminf_{N}E[f_{N}]\leq 4\liminf_{N}E[M_{N}^{2}].$$
Since the sequence $f_{N}$ is increasing, we can write the LHS as the possibly infinite $\sup_{t\geq 0}M_{t}^{2}$. For the RHS, we simply bound by the entire supremum
$$E[\sup_{t\geq 0}M_{t}^{2}]\leq 4\sup_{N\geq 0}E[M_{N}^{2}]<\infty.$$