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

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

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