Lemma 24.7.3.label Let $\mu: \cb_{\real^d}\to [0, \infty]$ be a regular Borel measure on $\real^{d}$ such that $\mu(\partial B_{\real^d}(x, r)) = 0$ for all $x \in \real^{d}$ and $r > 0$, then the average $(r, x) \mapsto A_{r}\mu(x)$ is jointly continuous.
Proof. Let $\seq{(r_n, x_n)}\subset (0, \infty) \times \real^{d}$ and $(r, x) \in (0, \infty) \times \real^{d}$ such that $(r_{n}, x_{n}) \to (r, x)$. Given that $\mu(\partial B_{\real^d}(y, s)) = 0$ for all $x \in \real^{d}$ and $s > 0$, $\one_{B_{\real^d}(x_n, r_n)}\to \one_{B_{\real^d}(x, r)}$ $\mu$-a.e. and Lebesgue-a.e. as $n \to \infty$.
Since $\seq{(r_n, x_n)}$ is convergent, there exists $K \subset \real^{d}$ compact such that $B_{\real^d}(x, r) \cup \bigcup_{n \in \natp}B_{\real^d}(x_{n}, r_{n}) \subset K$. By regularity of $\mu$, $\mu(K) < \infty$. Thus the Dominated Convergence Theorem implies that $A_{r_n}\mu(x_{n}) \to A_{r}\mu(x)$ as $n \to \infty$.$\square$
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