Sufficient conditions for the Hardy-Littlewood Maximal function $M(f)$ being continuous There are four common versions of Hardy-Littlewood Maximal operator $M(f)$: centered/uncentered + ball/cube.
I noticed that the continuity of $M(f)$ depends on the version. For example, let $f$ be the characteristic function of the interval $[0,1]$. Then the centered version of $M(f)$ is not continuous while the uncentered version is continuous. (ball and cube are the same in 1-dim). My question is: for each version of $M(f)$, under what conditions for $f$ is $M(f)$ continuous? 
 A: Here is a partial answer:
Let $Mf(x)=\sup_{r>0}\frac{1}{\lambda(B(x,r))}\int_{B(x,r)}|f(y)|dy$ be the (centered) Hardy Littlewood Maximal Function.

Proposition: Let $f\in L^1_{loc}(\mathbb{R^n})$. Then $Mf$ is lower semicontinuous, and in particular, measurable.

Proof: Lower semicontinuity means $E(t)=\{x: Mf(x)>t\}$ is open for all $t\in \mathbb{R}$. Let $x\in E(t)$. Then $Mf(x)>t$, so there is an $r>0$ such that $Mf(x)\geq\frac{1}{\lambda(B(x,r))}\int_{B(x,r)}|f(y)|dy>t$. If we choose $s>r$ small enough, then $$Mf(x)\geq\frac{1}{\lambda(B(x,r))}\int_{B(x,r)}|f(y)|dy>\frac{1}{\lambda(B(x,s))}\int_{B(x,r)}|f(y)|dy>t$$Then, for $z$ such that $|z-x|+r<s$, we will have $B(x,r)\subset B(z,s)$, so $$Mf(z)\geq\frac{1}{\lambda(B(z,s))}\int_{B(z,r)}|f(y)|dy>\frac{1}{\lambda(B(x,s))}\int_{B(x,r)}|f(y)|dy>t$$ Thus $z\in E(t)$, and the set of $z$ is open, so $E(t)$ is open.
Note: This is much easier to prove for the uncentered Hardy Littlewood Function.
Now, we are reduced to showing when $Mf$ is upper semicontinuous, since if a function is lower and upper semicontinuous somewhere, it is continuous. Searching hasn't gotten me much in terms of if and only if, but there is this result:

Proposition: If $f\in L^1_{loc}(\mathbb{R^n})$ is continuous at some point $x$, then $Mf$ is also continuous at $x$.

The proof is found in this article: Lemma 7.3. A slightly different requirement is the following:

Proposition: If $f\in L^1_{loc}(\mathbb{R^n})$ is upper semicontinuous at $x$ and $f(x)\leq Mf(x)$, then $Mf$ is continuous at $x$.

The proof is in this paper (Lemma 3.4). Although $\chi_{[0,1]}$ is upper semicontinuous everywhere, $M\chi_{[0,1]}$ is not continuous at $0$ or $1$. The theorem holds because $$\chi_{[0,1]}(0)=1>M\chi_{[0,1]}(0)=\frac{1}{2}$$This provides a counterexample showing that we need the second condition.
Hope this helps as a partial answer!
A: In Folland's Real Analysis : Modern Techniques and Their Applications, it is proved that $A_rf(x)$ is jointly continuous in $r$ and $x$. Hence, because the preimage of any open set is open under $A_rf$ and that you can write $Hf^{-1}(a,\infty)=\cup_{r>0} (A_r|f|)^{-1}(a,\infty)$, you have that $Hf$ is lower semicontinuous. Same applies to prove upper semicontinuity.
