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.
Best Answer
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:
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.