Why $\lim_{m\rightarrow\infty}F_m(x)=\liminf_{m\rightarrow\infty}F_m(x)=f(x)$?

Question

I am trying to understand a step in the proof of Use Fatou Lemma to show that $f$ takes real values almost everywhere.

it is shown that

$\lim_{m\rightarrow\infty}F_m(x)=\liminf_{m\rightarrow\infty}F_m(x)=f(x)$

Why the first equality is true?

See the proof:

Let $F_m(x)=\sum_{i=1}^m|f_{n+1}(x)-f_n(x)|$. Since $F_m(x)$ is increasing for each $x$, its limit in $m$ necessarily exists in the extended reals. So $\lim_{m\rightarrow\infty}F_m(x)=\liminf_{m\rightarrow\infty}F_m(x)=f(x)$

Then by Fatou:

$$\int_A|f(x)|^pd\mu\leq \liminf_m\int_A|F_m(x)|^p dx\leq\liminf_m\sum_{n=1}^m\frac{1}{2^{pn}}d\mu<\infty.$$

Taking $A$ to be $\mathbb{R}$, it follows that $\int_A|f(x)|^pd\mu<\infty$ which implies $\mu(\{|f|=\infty\})=0$.

Answer 1

$\lim a_n$ exists iff $\lim sup\, a_n=\lim inf \, a_n$ in which case $lim \, a_n=\lim sup \,a_n=\lim inf \, a_n$ . In this case $\lim a_n$ exists, so $lim \, a_n=\lim sup \, a_n=\lim inf \, a_n$ . (Infinite limits are allowed in this argument).

Answer 2

Whenever the limit exists (by monotonicity in your case) both $Liminf$ and $limsup$ exist and are equal to the limit.