The Lusin area function for a fixed admissible kernel and aperture

Assume Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Fix an aperture $a>0$ and an admissible kernel: a real-valued radial Schwartz function $\psi$ (Schwartz space and its seminorms) with ${\int }_{{\mathbb{R}}^n}\psi =0$ and $\psi$ not identically zero; the support convention of The support of a function on ${\mathbb{R}}^n$ and its compactly supported Riemann integral applies to the compactly supported functions used below.

For $t>0$ write ${\psi }_t(y):=t^{-n}\psi (y\mathrm{/}t)$, and for $x\in {\mathbb{R}}^n$ let ${\mathrm{\Gamma }}_a(x):=\{(y,t)\in {\mathbb{R}}^n\times (0,\mathrm{\infty }):\mathrm{\mid }y-x\mathrm{\mid }<at\}$ be the cone of aperture $a$ over $x$. For $f\in L^p({\mathbb{R}}^n;\mathbb{C})$ with $1\le p\le \mathrm{\infty }$, or for $f\in \mathcal{S}({\mathbb{R}}^n)$ (which lies in every $L^p$), fix a representative of $f$ and define $A_{a,\psi }f(x):=({\int }_0^{\mathrm{\infty }}{\int }_{\mathrm{\mid }y-x\mathrm{\mid }<at}\mid (f\ast {\psi }_t)(y){\mid }^{2}dy\frac{dt}{t^{n+1}}{)}^{1\mathrm{/}2}\in [0,\mathrm{\infty }],$ where $f\ast {\psi }_t$ is the convolution of Convolution of two functions on ${\mathbb{R}}^n$. The following well-definedness facts are part of the definition and are used with the cited suppliers.

1. Let $q$ be conjugate to $p$, with $q=\mathrm{\infty }$ when $p=1$ and $q=1$ when $p=\mathrm{\infty }$. For every $y\in {\mathbb{R}}^n$ and $t>0$, Holder's inequality (Complex Holder, Minkowski, and the quotient norm) makes $(f\ast {\psi }_t)(y):={\int }_{{\mathbb{R}}^n}f(u)t^{-n}\psi ((y-u)\mathrm{/}t)du$ absolutely convergent and independent of the representative of $f$. Young's inequality (Young's convolution inequality under Countable Choice) gives $\mathrm{\parallel }f\ast {\psi }_t{\mathrm{\parallel }}_p\le \mathrm{\parallel }\psi {\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_p$. Moreover $(y,t)\mapsto t^{-n}\psi ((y-\cdot )\mathrm{/}t)$ is continuous into $L^q({\mathbb{R}}^n)$ for $t>0$: near any fixed $(y,t)$ these Schwartz kernels depend pointwise continuously on the parameters and have a common integrable Schwartz majorant for finite $q$, so dominated convergence applies (Dominated convergence); for $q=\mathrm{\infty }$, uniform continuity on bounded sets and a uniform Schwartz tail give convergence in the supremum norm. Holder's inequality therefore shows that $(y,t)\mapsto (f\ast {\psi }_t)(y)$ is jointly continuous, hence Borel measurable.
2. For each $x$ the set ${\mathrm{\Gamma }}_a(x)$ is open in ${\mathbb{R}}^n\times (0,\mathrm{\infty })$ and the integrand is nonnegative, so the iterated integral over ${\mathrm{\Gamma }}_a(x)$ is well defined in $[0,\mathrm{\infty }]$ by Tonelli (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product) without any integrability hypothesis; the comparison with the integral over the measurable set ${\mathrm{\Gamma }}_a(x)$ is the convention of Integral over a measurable subset.
3. $A_{a,\psi }f$ is Borel measurable as a function of $x$, in fact it is lower semicontinuous. Write $H(y,t):=\mathrm{\mid }(f\ast {\psi }_t)(y){\mathrm{\mid }}^2$; this is continuous by item 1. If $x_m\to x$, then ${\mathbf{1}}_{\{\mathrm{\mid }y-x\mathrm{\mid }<at\}}\le {\mathrm{lim\hspace{0.17em}inf}}_m{\mathbf{1}}_{\{\mathrm{\mid }y-x_m\mathrm{\mid }<at\}}$ for every $(y,t)$, because the cone inequality is strict. Fatou's lemma (Fatou's lemma) applied to the nonnegative integrands with measure $dydt\mathrm{/}{t}^{n+1}$ gives $A_{a,\psi }f(x{)}^2\le {\mathrm{lim\hspace{0.17em}inf}}_m{A}_{a,\psi }f(x_m{)}^2$; hence the extended-valued function $A_{a,\psi }f$ is lower semicontinuous and therefore Borel.
4. If two representatives of $f$ agree almost everywhere, their integrands in the integral formula of item 1 agree almost everywhere in $u$; the Lebesgue integral respects almost-everywhere equality (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree), so the convolutions, and hence the area functions, agree at every $(y,t)$ and $x$. Thus the functional is defined on almost-everywhere classes, with values allowed to equal $+\mathrm{\infty }$.

The cancellation $\int \psi =0$ and the aperture $a$ are part of the data. Distinct pairs need not give distinct functionals: replacing $\psi$ by $-\psi$ leaves $A_{a,\psi }$ unchanged, since the squared modulus of every convolution is unchanged. No equivalence between this functional and a square-function scale is asserted here.