Show that $Mf_n(x) \nearrow Mf(x)$, where $f_n(x) \nearrow f(x)$ almost everywhere 
I am trying to show that $Mf_n(x) \nearrow Mf(x)$, where $f_n(x) \nearrow f(x)$ almost everywhere $x$, $f_n(x)$ and $f(x)$ are nonnegative and locally integrable, and
$$ Mf(x) = \sup_{r > 0}\frac{1}{|B(x,r)|}\int_{B(x,r)} |f(y)|dy$$
is the Hardy-Littlewood maximal function.

My question involves the requirements to interchange the $\limsup$, i.e.
$$
\lim_{n \to \infty} Mf_n(x) = \lim_{n \to \infty}\left(\sup_{r > 0}\frac{1}{|B(x,r)|}\int_{B(x,r)} |f_n(y)|dy\right) = \sup_{r > 0}\frac{1}{|B(x,r)|}\left(\lim_{n \to \infty}\int_{B(x,r)} |f_n(y)|dy\right).$$
If one can do this then the rest follows easily enough via the MCT and nonnegativity of $f_n(x)$. Is it enough to interchange if $f_n(x)$ monotone increasing, pointwise convergent, and locally integrable? If so, why as it isn't quite obvious to me?
 A: Let $\left( f_n \right)_{n \in \mathbb{N}}$ be an increasing sequence of nonnegative locally integrable functions. Let $f=\lim_n f_n$. Given $n \in \mathbb{N}$, let $y_n=\sup_{r>0} \frac{1}{m \left( B(x,r) \right)} \int_{B(x,r)} f_n d m$.
It's easy to see that given
$n \in \mathbb{N}$, $$y_n=\sup_{r>0} \frac{1}{m\left(B(x,r)\right)} \int_{B(x,r)} f_n dm \leq \sup_{r>0} \frac{1}{m\left(B(x,r)\right)} \int_{B(x,r)} f dm.$$ Indeed: this inequality is obvious without the suprema in both sides for any fixed $r>0$. In particular, it also holds that
$$\lim_n y_n \leq \sup_{r>0} \frac{1}{m\left(B(x,r)\right)} \int_{B(x,r)} f dm.$$
The reverse inequality is somewhat tricky. We claim that $\left( y_n \right)_{n \in \mathbb{N}}$ is an increasing sequence. Indeed: fix $n \in \mathbb{N}$. $f_n \leq f_{n+1}$, therefore, for any $r>0$:
$$
\frac{1}{m \left( B(x,r) \right)} \int_{B(x,r)} f_n d m \leq \frac{1}{m \left( B(x,r) \right)} \int_{B(x,r)} f_{n+1} d m \leq y_{n+1}.
$$
$r>0$ is arbitrary, so $y_n \leq y_{n+1}$.
Any increasing sequence admits a limit and limits preserve inequalities. Threfore, given $r>0$:
$$
\lim_n y_n \geq \lim_n \frac{1}{m \left( B(x,r) \right)} \int_{B(x,r)} f_n d m=\frac{1}{m \left( B(x,r) \right)} \int_{B(x,r)} f d m
$$
$r>0$ is arbitrary, so
$$
\sup_{r>0} \frac{1}{m \left( B(x,r) \right)} \int_{B(x,r)} f d m
\leq
\lim_n y_n.
$$
A: These limits are interchangeable when you have monotone convergence:
One inequality is obvious. For the other direction let $c< Mf(x)$. Then there is $r_c>0$ so that $M_{r_c} f(x)>c$ (in the obvious notation when taking a ball of fixed radius). Now, $M_{r_c} f_n(x) \rightarrow M_{r_c} f(x)$ by the MCT, so $\lim_n M f_n(x)> c$ as well and the result now follows.
