Almost sure and order convergence are equivalent in $L_p$ spaces

Question

I have attempted to prove the following lemma, which is given as an exercise by Aliprantis and Border. Is the proof correct? Improvements are welcome.

Lemma. An order bounded sequence $(f_n)$ in some $L_p(\mu)$ space satisfies $f_n \xrightarrow{o} f$ if and only if $f_n \to f$ a.s. ($\mu$). (The symbol $\xrightarrow{o}$ denotes order convergence.)

Proof. First note that since $(f_n)$ is order bounded $|f_n - f| \leq h + |f|$ holds for some $h \in L_p(\mu)$ and all $n$. Therefore, $\sup_{m \geq n}|f_n - f| \in L_p(\mu)$ for all $n$.

Assume $f_n \xrightarrow{o} f$, which means that there exists a sequence $(g_n) \in L_p(\mu)$ such that $|f_n - f| \leq g_n \downarrow 0$. The inequality means that $|f_n(x) - f(x)| \leq g_n(x)$ for $\mu$-almost every $x$ and all $n$. Thus, $\limsup_n |f_n(x) - f(x)| \leq \limsup_n g_n(x)=0$ for $\mu$-almost every $x$. This implies $f_n \to f$ a.s. ($\mu$).

Now assume $f_n \to f$ a.s. ($\mu$). Let $g_n := \sup_{m \geq n}|f_m - f|$. Then, $|f_n - f| \leq g_n \downarrow 0$.

Answer 1

The proof looks correct, up to two small typos:

• $\limsup_n |f_m(x) - f(x)| \leq \limsup_n g_n(x)=0$ should be $\limsup_n |f_n(x) - f(x)| \leq \limsup_n g_n(x)=0$;
• in the last line, the definition of $g_n$ should be $g_n := \sup_{m \geq n}|f_m - f|$.