Exercise 1.6.22 (Two-sided Hardy-Littlewood inequality II)

Question

Let $\mu$ be a Borel measure (i.e. a countably additive measure on the Borel $\sigma$-algebra) on ${\bf R}$, such that $0 < \mu(I) < \infty$ for every interval $I$ of positive length. Assume that $\mu$ is inner regular, in the sense that $\mu(E) = \sup_{K \subset E, \hbox{ compact}} \mu(K)$ for every Borel measurable set $E$. Establish the Hardy-Littlewood maximal inequality

$$\displaystyle \mu \left( \left\{ x \in {\bf R}: \sup_{x \in I} \frac{1}{\mu(I)} \int_{I} |f| \ d\mu \geq \lambda \right\} \right) \leq \frac{2}{\lambda} \int_{\bf R} |f|\ d\mu$$

for any absolutely integrable function $f \in L^1(\mu)$, where the supremum ranges over all open intervals $I$ that contain $x$. Note that this essentially generalises \href{./exercise_1-6-11}{Exercise 6.11}, in which $\mu$ is replaced by Lebesgue measure.

Elu Thingol · Accepted Answer

It suffices to verify the claim with strict inequality,
$$\displaystyle \mu \left( \left\{ x \in {\bf R}: \sup_{x \in I} \frac{1}{\mu(I)} \int_{I} |f| \ d\mu > \lambda \right\} \right) \leq \frac{2}{\lambda} \int_{\bf R} |f|\ d\mu,$$
as the non-strict case then follows by perturbing $\lambda$ slightly and then taking limits.\par

Let $\mu$ be an arbitrary inner regular measure on the Borel space $(\mathbf{R},\mathcal{B})$. Fix $f$ and $\lambda$. By inner regularity, it suffices to show that
$$\mu\left(K\right) \leq \frac{2}{\lambda} \int_{\bf R} |f|\ d\mu$$
whenever $K$ is a compact set that is contained in
$$\left\{ x \in {\bf R}: \sup_{x \in I} \frac{1}{\mu(I)} \int_{I} |f| \ d\mu > \lambda \right\}.$$

By construction, for every $x \in K$, there exists an open interval $I_x$ such that
\begin{equation}
\frac{1}{\mu(I)} \int_{I} |f| \ d\mu > \lambda.
\label{eq:1.28}
\end{equation}

By compactness of $K$, we can cover $K$ by a finite number $I_1,\ldots,I_n$ of such open intervals. Applying the \href{./exercise_1-6-21}{Besicovitch covering lemma in one dimension}, we can find a subcollection $I_{1}',\ldots,I_{m}'$ of intervals such that
\begin{enumerate}[label=(\roman*)]
\item $\bigcup_{i=1}^n I_n = \bigcup_{j=1}^m I'_m;$ and
\item Each point $x \in {\bf R}$ is contained in at most two of the $I'_m$.
\end{enumerate}

In other words, we have
$$\mu\left(\bigcup_{i=1}^{n} I_i \right) \leq 2 \cdot \sum_{j=1}^{m} \mu(I_j').$$

By (\ref{eq:1.28}), on each interval $I_j'$ we have
$$\mu\left(I_j'\right) < \frac{1}{\lambda} \int_{I_j'} |f| \ d\mu;$$
summing in $j$ and using the property (ii) of the Besicovitch covering lemma, we conclude that
$$\mu\left(\bigcup_{i=1}^{n} I_i \right) \leq \frac{2}{\lambda} \int_{\mathbf{R}} |f| \ d\mu.$$
Since the intervals $I_1,\ldots,I_n$ cover $K$, we obtain the desired assertion.

Exercise 1.6.22 (Two-sided Hardy-Littlewood inequality II)

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Exercise 1.6.22 (Two-sided Hardy-Littlewood inequality II)

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