The weighted maximal function of a doubling weight

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)).

Let $v$ be a weight (Weights, their associated measures, and the spaces L^p(w)) whose measure $vd\lambda$ is doubling with constant $c_v$: $v(B(x,2r))\le c_{v}v(B(x,r))(x\in {\mathbb{R}}^n,r>0)\mathrm{.}$

For $f\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(v)$ the weighted maximal function is $M^{v}f(x):={\sup }_{r>0}\frac{1}{v(B(x,r))}{\int }_{B(x,r)}\mathrm{\mid }f\mathrm{\mid }vd\lambda ,$ and its cube analogue is $M_c^{v}f(x):={\sup }_{Q\ni x}\frac{1}{v(Q)}{\int }_Q\mathrm{\mid }f\mathrm{\mid }vd\lambda ,$ the supremum over the axis-parallel cubes of Axis-parallel cubes, their averages, and cube maximal functions containing $x$. The weighted averages are finite because $f\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(v)$ and $0<v(Q)<\mathrm{\infty }$ for bounded $Q$.

Ball-cube comparability. Balls and cubes of comparable size have $v$-measure comparable by a constant depending only on $n$ and $c_v$: for a cube $Q$ with centre $y$ and side length $2r$ one has $Q\subseteq B(y,\sqrt{n}r)\subseteq Q(y,\sqrt{n}r)$, and iterating the doubling inequality a number of times depending only on $n$ gives $v(Q)\le v(B(y,\sqrt{n}r))\le C(n,c_v)v(B(y,r))\le C(n,c_v)v(Q)$, while $B(x,r)\subseteq Q(x,r)\subseteq B(x,\sqrt{n}r)$ gives the same comparison for balls. Consequently $M^{v}f\le C(n,c_v)M_c^{v}f$ and $M_c^{v}f\le C(n,c_v)M^{v}f$ pointwise: a cube $Q=Q(y,r)\ni x$ is contained in $B(x,2\sqrt{n}r)\subseteq B(y,3\sqrt{n}r)$, whose $v$-measure is at most a dimensional number of doublings times $v(B(y,r))\le v(Q)$; conversely each centred ball $B(x,r)$ lies in $Q(x,r)$ of comparable $v$-measure. These containments compare each average to one in the appropriate supremum (Ball and cube maximal functions are pointwise comparable provides the unweighted geometric sandwich, and the doubling of $vd\lambda$ converts it to a $v$-measure comparison).

Measurability. For each fixed $r>0$ the function $x\mapsto v(B(x,r){)}^{-1}{\int }_{B(x,r)}\mathrm{\mid }f\mathrm{\mid }vd\lambda$ is continuous: the numerator is continuous in the centre by dominated convergence with dominating function $\mathrm{\mid }f\mathrm{\mid }v$ over a fixed bounded ball containing all the translates, and the denominator $x\mapsto v(B(x,r))$ is continuous by dominated convergence with dominating function $v$ over such a ball; the denominator is positive (Dominated convergence, and Ball averages vary continuously with the centre and radius for the unweighted averages that underlie the same argument). The same dominated-convergence argument gives continuity in $r$, so the supremum over positive radii equals that over positive rational radii. Hence $M^{v}f$ is Borel measurable (Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable); the same holds for $M_c^{v}f$.