Ball averages vary continuously with the centre and radius

Choose $\rho >0$ with $0<\rho <r\mathrm{/}2$. For all sufficiently large $k$, [given, choose, algebra] $\mathrm{\mid }{r}_k-r\mathrm{\mid }<\rho$ and $\mathrm{\parallel }{x}_k-x{\mathrm{\parallel }}_2<\rho$. Then $B(x,r-\rho )\subseteq B(x_k,r_k)\subseteq B(x,r+2\rho )\mathrm{.}$ Since $f$ is locally integrable, the function $g:=\mathrm{\mid }f\mathrm{\mid }{\mathbf{1}}_{B(x,r+2\rho )}$ is integrable.

Put ${\chi }_k:={\mathbf{1}}_{B(x_k,r_k)}$ and $\chi :={\mathbf{1}}_{B(x,r)}$. [algebra] If $y\mathrm{\notin }\mathrm{\partial }B(x,r)$, then $\mathrm{\parallel }y-x{\mathrm{\parallel }}_2\mathrm{\ne }r$, so for all sufficiently large $k$ the membership of $y$ in $B(x_k,r_k)$ agrees with its membership in $B(x,r)$. Thus ${\chi }_k(y)\to \chi (y)$ for every $y\mathrm{\notin }\mathrm{\partial }B(x,r)$.

The boundary sphere satisfies [L2, L3, algebra] $\mathrm{\partial }B(x,r)\subseteq B(x,r+\frac{1}{m})\setminus B(x,r-\frac{1}{m})(m>1\mathrm{/}r)\mathrm{.}$ By [L2] and [L3], $\lambda \text{\hspace{0.17em}⁣}(B(x,r+\frac{1}{m})\setminus B(x,r-\frac{1}{m}))=\lambda (B(0,1))((r+\frac{1}{m}{)}^n-(r-\frac{1}{m}{)}^n)\to 0,$ so $\lambda (\mathrm{\partial }B(x,r))=0$.

By [L2] and [L3], [L2, L3, algebra] $\lambda (B(x_k,r_k))=\lambda (B(0,1))r_k^n\to \lambda (B(0,1))r^n=\lambda (B(x,r))\mathrm{.}$ The limit denominator is positive by [L2].

Steps 1.1, 1.2, and 1.3 let us apply [L4] to [step 1.1, step 1.2, step 1.3, L4] $f{\chi }_k$, dominated by $g$, and obtain $\int f(y){\chi }_k(y)d\lambda (y)\to \int f(y)\chi (y)d\lambda (y)\mathrm{.}$ In other words, ${\int }_{B(x_k,r_k)}fd\lambda \to {\int }_{B(x,r)}fd\lambda \mathrm{.}$

Combining steps 2.1 and 1.4 yields [L1, step 2.1, step 1.4, algebra] $A_{r_k}f(x_k)\to A_{r}f(x)\mathrm{.}$ Since the approximating sequence was arbitrary, $\mathrm{\Phi }$ is continuous.