The Hardy–Littlewood maximal function is Borel measurable

Question

Let $f : \mathbb R^n \to \mathbb R$ be Lebesgue-integrable. We define the Hardy–Littlewood maximal function $Mf: \mathbb R^n \to [0, \infty]$ by

$$ M f(x) :=\sup _{r>0} \frac{1}{|B(x, r)|} \int_{B(x, r)} |f(t)| \mathrm d t $$ where $B(x, r)$ is the open ball of radius $r$ centred at $x$, and $|E|$ denotes the $n$-dimensional Lebesgue measure of $E \subset \mathbb R^n$.

Theorem: $Mf$ is Borel measurable.

I'm trying to adapt the strategy in this thread to higher dimension. Could you have a check on my attempt?

Proof: Let $$ g_r (x) := \frac{1}{|B(x, r)|} \quad \text{and} \quad h_r(x) := \int_{B(x, r)} |f(t)| \mathrm d t \quad \forall r>0, \forall x \in \mathbb R^n. $$

Because Lebesgue measure is translation-invariant, we get $g_r (x) = g_r (y)$ for all $x, y \in \mathbb R^n$. This means $g_r$ is constant. Let's prove that $h_r$ is continuous. Let $x, x_n \in \mathbb R^n$ such that $x_n \to x$. Because the $n$-dimensional Lebesgue measure of a sphere is $0$, we have $1_{B(x_n, r)} \to 1_{B(x, r)}$ a.e. It follows that $$ 1_{B(x_n, r)} |f| \to 1_{B(x, r)} |f| \quad \text{a.e.} $$

By dominated convergence theorem, we get $h_r (x_n) \to h_r (x)$ and thus $(g_rh_r) (x_n) \to (g_rh_r) (x)$. It follows that $g_rh_r$ is measurable. The supremum of a collection of measurable functions is measurable, so $$ Mf = \sup_{r>0} g_rh_r $$ is measurable.

Answer 1

Here is a little refinement of your sketch of proof.

A function $f$ on $\mathbb{R}^n$ is called locally integrable, denoted $f\in L^1_{loc}(\mathbb{R}^d,\lambda_d)$, if $f\mathbb{1}_E\in\mathcal{L}_1(\mathbb{R}^d,\lambda_d)$ for all bounded measurable set $E$.

Claim: For such function $f$ and fixed $r>0$, the map $$M_rf:x\mapsto \frac{1}{\lambda(B(x;r))}\int_{B(x;r)}f\,d\lambda=\frac{1}{\omega_n r^n}\int_{B(x;r)}f\,d\lambda$$ is continuous and hence measurable.

Indeed, for fixed $x\in\mathbb{R}^d$ we have $f\mathbb{1}_{B(x;2r)}\in L^1$. Since $\|\mathbb{1}_{B(y;r)}-\mathbb{1}_{B(x;r)}\|_1\xrightarrow{y\rightarrow x}0$, the conclsion follows from \begin{align*} \Big|\int_{B(y;r)}f\,d\lambda-\int_{B(x;r)}f\,d\lambda\Big|\leq \int_{B(x;r)\triangle B(y;r)}|f\mathbb{1}_{B(x;2r)}|\,d\lambda. \end{align*} for $|x-y|<r$. Continuity follows from the following basic result:

$f\in L_1(\mu)$ then for any $\varepsilon>0$, there is $\delta>0$ such that $\mu(A)<\delta$ implies that $\int_A|f|\,d\mu<\varepsilon$.

It follows that the map $x\mapsto\sup_{r>0}\frac{1}{\lambda(B(x;r))}\int_{B(x;r)}\,d\lambda=\sup_{r>0}M_rf(x)$ is lower semicontinuous and so, measurable.

Answer 2

Here's another elementary approach $:$

Given $\alpha \in \mathbb R,$ consider the set $A_{\alpha} : = \left \{x \in \mathbb R^n\ |\ Mf(x) \gt \alpha \right \}.$ We claim that $A_{\alpha}$ is open and hence Borel measurable in particular.

Take $x \in A_{\alpha}$ arbitrarily. Then $M f(x) \gt \alpha$ i.e.

$$\sup\limits_{r \gt 0} \frac {1} {\left \lvert B(x,r) \right \rvert} \int_{B(x,r)} |f(t)|\ dt > \alpha.$$ So by the property of supremum there exists $r_0 \gt 0$ such that $$\frac {1} {\left \lvert B(x,r_0) \right \rvert} \int_{B(x,r_0)} |f(t)|\ dt > \alpha.$$ Choose $r' \gt r \gt 0$ such that $$\tag{*}\frac {1} {\left \lvert B(x,r') \right \rvert} \int_{B(x,r_0)} |f(t)|\ dt > \alpha.$$ This is possible by the virtue of intermediate value theorem as the function $x \mapsto \frac {1} {x^n}$ is continuous and decreasing on $(0, \infty)$ and since $$\left \lvert B(y,s) \right \rvert = \left \lvert B(0,s) \right \rvert = c_n s^n$$ for any $y \in \mathbb R^n$ and $s \gt 0$ (by the translation invariance of the Lebesgue measure), where $c_n = \left \lvert B(0,1) \right \rvert$ is an absolute constant (independent of $y$ and $s$). Now let $x' \in B(x,r' - r_0)$ so that $\|x' - x\| \lt r' - r_0.$

Claim $:$ $B(x,r_0) \subseteq B(x',r').$

Indeed if $y \in B(x,r_0)$ then $\|y - x\| \lt r_0.$ Hence it follows that $$\|y - x'\| \leq \|y - x\| + \|x - x'\| \lt r_0 + (r' - r_0) = r'$$ i.e. $y \in B(x',r'),$ as claimed. So we have by $(*)$ $$\alpha \lt \frac {1} {\left \lvert B(x,r') \right \rvert} \int_{B(x,r_0)} |f(t)|\ dt \leq \frac {1} {\left \lvert B(x',r') \right \rvert} \int_{B(x',r')} |f(t)|\ dt$$ Here $\left \lvert B(x,r') \right \rvert = \left \lvert B(x',r') \right \rvert,$ by the translation invariance of the Lebesgue measure. This shows that $$\sup\limits_{r \gt 0} \frac {1} {\left \lvert B(x',r) \right \rvert} \int_{B(x',r)} |f(t)|\ dt \gt \alpha$$ i.e. $M f(x') \gt \alpha$ and hence $x' \in A_{\alpha}.$ So it follows that $B(x,r' - r_0) \subseteq A_{\alpha},$ which shows that $A_{\alpha}$ is an open set. $\blacksquare$