Why doe it use limsup?(Real analysis by Folland) 
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*In the proof of the theorem, it defines 
$$F_{k}=\{x \in A :\limsup_{r\to 0} \frac{\lambda(B(r,x))}{m(B(r,x))}>\frac{1}{k} \}$$ and it proves that $m(F_{k})=0$.
Why did not it define $F_{k}$ as follows:
$$F_{k}=\{x \in A :\lim_{r\to 0} \frac{\lambda(B(r,x))}{m(B(r,x))}>\frac{1}{k} \}$$

*In theorem 3.20 after reaching to this point that
$$\frac{1}{m(B(r,x))}\int_{m(B(r,x))}|f(y)-f(x)|dy \le\frac{1}{m(B(r,x))}\int_{m(B(r,x))}|f(y)-c|+\epsilon$$
then it uses limsup as follows:
$$\limsup_{r\to 0}\frac{1}{m(B(r,x))}\int_{m(B(r,x))}|f(y)-f(x)|dy \le|f(x)-c|+\epsilon$$
why did not it use lim as follows:
$$\lim_{r\to 0}\frac{1}{m(B(r,x))}\int_{m(B(r,x))}|f(y)-f(x)|dy \le\lim_{r\to 0}\frac{1}{m(B(r,x))}\int_{m(B(r,x))}|f(y)-c|+\epsilon=|f(x)-c|+\epsilon$$
 A: Not sure if this will help but I just wanted to show my proof.
For each $c\in \mathbb{C}$ we can apply Theorem 3.18 to $g_c(x) = |f(x) - c|$ and see that outside a set $E_c$ with $m(E_c) = 0$ we have $$\lim_{r\rightarrow 0}\frac{1}{m(B(r,x))}\int_{B(r,x)}|f(y) - c| dy = |f(x) - c|$$
Let $D$ be a countable dense subset in $\mathbb{C}$ and $E = \bigcup_{c\in D}E_c$ so $m(E) = 0$. Now if $x\notin E$ then given $\epsilon >0$, choose a $c\in D$ with $|f(x) - c| < \epsilon/2$. Then
\begin{align*}
\lim_{r\rightarrow 0}\frac{1}{m(B(r,x)}\int_{B(r,x)}|f(y) - f(x)|dy &\leq \lim_{r\rightarrow 0}\frac{1}{m(B(r,x)}\int_{B(r,x)}|f(y) - c| + |c - f(x)|dy\\
&= \left(\lim_{r\rightarrow 0}\frac{1}{m(B(r,x)}\int_{B(r,x)}|f(y) - c|dy\right) + |f(x) - c|\\
&= |f(x) - c| + |f(x) - c|\\
&< \epsilon/2 + \epsilon/2\\
&= \epsilon
\end{align*}
(We need $D$ to be countable in order that we can ensure that  $E = \bigcup_{c\in D}E_c$ is a countable union of null sets and so it is a null set. We need $D$ to be dense to be able to use it to estimate  $\lim_{r\rightarrow 0}\frac{1}{m(B(r,x)}\int_{B(r,x)}|f(y) - f(x)|dy$.)
