Assumptions on the Borel measure in Stein's Harmonic Analysis

Question

I am currently reading the proof of Theorem 1 on Page 13 of Stein's Harmonic Analysis which proves that if $f \in L^{1}(\mathbb{R}^{n})$, then for every $\alpha > 0$, $$\mu(\{x: (Mf)(x) > \alpha\}) \leq \frac{c_{2}}{\alpha}\int_{\mathbb{R}^{n}}|f(y)|\, d\mu(y)$$ where $$(Mf)(x) = \sup_{\delta > 0}\frac{1}{\mu(B(x, \delta))}\int_{B(x, \delta)}|f(y)|\, d\mu(y).$$

The strategy of course is to take a compact subset of $\{x: (Mf)(x) > \alpha\}$ and work with this compact subset instead.

This proof seems to implicitly imply that $\mu$ is inner regular. However, I do not see where this is stated in the book. Stein's assumptions on $\mu$ are as follows:

1. $\mu$ is a nonnegative Borel measure such that $\mu(\mathbb{R}^{n}) > 0$.
2. $\mu$ is a doubling measure, that is, there exists constants $c_{1}, c_{2}$ such that for all $x, y, \delta$, $\mu(B(x, c_{1}\delta)) \leq c_{2}\mu(B(x, \delta))$.
3. For each open set $U$ and each $\delta > 0$, the function $x \mapsto \mu(B(x, \delta) \cap U)$ is continuous.

Are these assumptions on $\mu$ strong enough to imply that $\mu$ is inner regular? Am I missing an additional assumption on $\mu$?

Answer 1

Stein also assumes on pg. 8 that $B(x,\delta_{1})\subset B(x,\delta_{2})$ when $\delta_{1}<\delta_{2}$, $$B(x,\delta)\cap B(y,\delta)\neq\emptyset\Longrightarrow B(y,\delta)\subset B(x,c_{1}\delta),$$ and $$\mu(B(x,c_{1}\delta))\leq c_{2}\mu(B(x,\delta))$$

All of the hypotheses on $\mu$ are enough to guarantee that there exists a ball $B=B(x,\delta)$ such that $0<\mu(B)<\infty$, although I don't think this is all that obvious. From this it follows that $0<\mu(B)<\infty$ for all balls $B$ (see below), and since $\bigcup_{\delta>0}B(x,\delta)=\mathbb{R}^{n}$, for any $x\in \mathbb{R}^{n}$ fixed, $\mu$ is a $\sigma$-finite Borel measure. One can then show that any $\sigma$-finite Borel measure on $\mathbb{R}^{n}$ is regular; I leave this last part as an exercise to you.

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Assume for the sake of a contradiction that $\mu(B)=\infty$ for every ball $B$.

Lemma 1. For all $x\in\mathbb{R}^{n}$ and $R>0$, there exists $\delta_{0}>0$ such that $\overline{B}(x,\delta_{0})\subset B_{R}(x)$, where $B_{R}(0)$ denotes the Euclidean ball centered at $x$ of radius $R$.

Proof. Suppose no such $\delta_{0}>0$ exists. Then there exists a sequence $\delta_{n}\downarrow 0$ such that $\overline{B}(x,\delta_{n})\cap B_{R}(x)^{c}\neq\emptyset$ and we can extract a sequence $y_{n}\in\overline{B}(x,\delta_{n})$ with $\left|y_{n}-x\right|\geq R$ and $y_{n}\in\overline{B}(x,\delta_{1})$ for all $n$ by monotonicity.

I claim that $y_{n}\rightarrow x$. Since $\left\{y_{n}\right\}$ is a sequence in a closed bounded set $\overline{B}(x,\delta_{1})$, it has a subsequence, which we also denote by $\left\{y_{n}\right\}$, converging to some $y\in\overline{B}(x,\delta_{1})$ by the Heine-Borel theorem. I claim that $y\in\overline{B}(x,\delta_{n})$ for all $n$. Indeed, if we delete the first $m$ terms of the sequence $\left\{y_{n}\right\}$, we obtain a subsequence containing $\overline{B}(x,\delta_{m})$. Applying the Heine-Borel theorem and using the uniqueness of subsequential limits, we see that $y\in\overline{B}(x,\delta_{m})$. Since $\bigcap_{\delta>0}\overline{B}(x,\delta)=\left\{x\right\}$, we conclude that $y=x$. But $$0=\left|y-x\right|=\lim_{n\rightarrow\infty}\left|y_{n}-x\right|\geq R>0,$$ which is a contradiction. $\Box$

As a consequence of the lemma, we obtain that $\mu$ is infinite on every Euclidean ball and therefore on every open set.

We can use the lemma to obtain a contradiction from the continuity of the function $x\mapsto \mu(B(x,\delta)\cap U)$ for a fixed open set $U$ and $\delta>0$. Fix $y\in\mathbb{R}^{n}$, and let $\delta>0$ be such that $\overline{B}(y,c_{1}\delta)\subset B_{1}(0)$, and take $U=B(y,\delta)$ . If $B(ty,\delta)\cap B(y,\delta)\neq\emptyset$ for all $t\geq 1$, then $$B(ty,\delta)\subset B(y,c_{1}\delta)\subset B_{1}(0),\quad\forall t\geq 1$$ which is a contradiction, as we can let $\left|ty\right|\rightarrow\infty$. So there exists some $t_{0}\geq 1$ such that $$B(ty,\delta)\cap B(y,\delta)=\emptyset$$ Since the function $$\varphi(t):=\mu(B(ty,\delta)\cap B(y,\delta)), \quad t>0$$ is continuous by hypothesis and $\varphi(1)=\mu(B(y,\delta))=\infty$. By the intermediate value theorem, there exists some $1<t_{1}<t_{0}$, such that $0<\varphi(t_{1})<\infty$. But then $B(t_{1}y,\delta)\cap B(y,\delta)$ is a nonempty open set and as observed above, we must have that $\varphi(t_{1})=\infty$, which is a contradiction.

If $0<\mu(B)<\infty$, for some ball $B=B(x,\delta)$, then I claim that $0<\mu(B(y,\delta))<\infty$ for every $y\in\mathbb{R}^{n}$ and $\delta>0$. Since $B(x,\delta)\subset B(x,\delta')$, for any $\delta'>\delta$, we have that $$\bigcup_{n=1}^{\infty}B(x,c_{1}^{n}\delta)=\bigcup_{\delta'>0}B(x,\delta')=\mathbb{R}^{n}$$ By using the monotonicity of the balls and iterating the doubling condition, we have that $$0<\mu(B(x,\delta))\leq\mu(B(x,c_{1}^{n}\delta)\leq c_{2}^{n}\mu(B(x,\delta))<\infty,\quad\forall n$$ In particular, $\mu$ is a $\sigma$-finite Borel measure. Since $B(x,c_{1}^{n}\delta)\cap B(y,\delta')\neq\emptyset$ for all $n$ sufficiently large, we have that $B(y,\delta')\subset B(x,c_{1}^{n}\delta)$ for all $n$ sufficiently large. Hence, $$\mu(B(y,\delta'))\leq\mu(B(x,c_{1}^{n}\delta))<\infty,\quad\text{all $n$ sufficiently large}$$

Inner regularity is then a consequence of the following more general result:

Lemma 2. Every $\sigma$-finite Borel measure on $\mathbb{R}^{n}$ is regular.