Show that $Mf_n(x) \nearrow Mf(x)$, where $f_n(x) \nearrow f(x)$ almost everywhere

Question

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?

Answer 1

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 $$ \sup_{r>0} \frac{1}{m \left( B(x,r) \right)} \int_{B(x,r)} f d m \leq \lim_n y_n. $$

Answer 2

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.