I am trying to understand a step in the proof of [Use Fatou Lemma to show that $f$ takes real values almost everywhere.](https://math.stackexchange.com/questions/3270130/use-fatou-lemma-to-show-that-f-takes-real-values-almost-everywhere/3270173#3270173)

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 $\mu(\{|f|=\infty\})=0$.