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I'm studying on 'Real Analysis' of M.Stein & R. Shakarchi, in particular i have a question about the proof propused about the weak-type inequality for maximal function (pag 102-104). I briefly summarize the reasoning:

Def(1): Let f an integrable function on $\mathbb{R}^d$, we define its maximal function by $$f^*(x) = \sup_{x\in B}\frac{1}{m(B)}\int_B |f(y)| dy, \hspace{0.25cm} x \in \mathbb{R}^d$$ where the supremum is taken over all balls containing the point x, and m is the Lebesgue measure.

Lemma(1): Let $\mathscr{B}=\{B_1, B_2, .., B_N\}$ a finite collection of open balls of $\, \mathbb{R}^d$, Then there exist a disjoint sub-collection $\{B_{i_{j}}\}_{j=1}^k$ such that: $$m(\cup_{\ell=1}^N B_\ell) \leq 3^d \sum_{j=1}^k m(B_{i_{j}})$$

Thm(weak-type inequality for maximal function): Let f an integrable function on $\mathbb{R^d}$, Let $E_{\alpha} = \{x \in \mathbb{R^d} | f^*(x) > \alpha\}$, Then $$m(E_\alpha) \leq \frac{3^d}{\alpha} ||f||_{L^1(\mathbb{R^d})}$$ Proof of Thm:

  1. It's easy to see that $E_\alpha$ is an open set, so in particular: $\forall x\in E_{\alpha} \,\exists B_x, \text{open ball}, x \in B_x : \frac{1}{m(B_x)}\int_{B_x} |f(y)| dy$, in other view: $$\text{(A)} \hspace{1cm}m(B_x) < \frac{1}{\alpha} \int_{B_{x}}|f(y)|dy$$

  2. We take $K \subset E_{\alpha}$ compact, surely $K \subset \cup_{x\in E_{\alpha}} B_x$, since compactness of $K$ there exists a finite sub-cover $K \subset \cup_{i=1}^N B_i$.

  3. We use the Lemma(1) on this sub-cover and obtain a disjoint sub-collection $\{B_{i_j}\}_{j=1}^M$.

  4. Therefore $m(K) \leq m(\cup_{i=1}^N B_i) \overset{Lemma(1)}{\leq} 3^d \sum_{j=1}^M m(B_{i_j}) \overset{(A)}{\leq} \frac{3^d}{\alpha} \sum_{j=1}^M\int_{B_{i_j}}|f(y)|dy = \frac{3^d}{\alpha} \int_{\cup_{j=1}^M B_{i_j}}|f(y)|dy \leq \frac{3^d}{\alpha} \int_{\mathbb{R^d}}|f(y)|dy = \frac{3^d}{\alpha} ||f||_{L^1(\mathbb{R^d})}$

  5. Since this inequality is true for all compact subset of $E_{\alpha}$ the proof is complete $\blacksquare$

Now my question: I cannot understand the 5. step, in particular, i would agree if $m(E_\alpha)< \infty$, because in these condition i can approximate every measurable set with a compact set. I tryed to proof the finite measure of $E_\alpha$ but without success, i don't know if maybe the fact that $E_\alpha$ is open as well as measurable can help me.

Manuel Bonanno
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1 Answers1

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By the Lebesgue monotone convergence theorem, $m(E_\alpha)=\lim_{n\to\infty}m(E_\alpha\cap B(0,n))$. For the same reason, every measurable set can be approximated from the inside with compact sets ( in a measure sense).

Gio67
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  • I understand the fist step of you're resoning, however I am not convinced of the second one. In particular: Let be $E_{\alpha}$ a measurable set with $m(E_{\alpha})=\infty$, let $K \subset E_{\alpha}$ compact. Because $K \subset \mathbb{R^d}$ since thm(Heine-Borel) K is closed and bounded, so $m(K)< C \in \mathbb{R}-{\infty}$. Assuming by absurd that $K$ approx $E_{\alpha}$ from inside, in other world assuming that $m(E-K)\leq \epsilon$, then by countable sub-additivity $m(E_{\alpha})=m((E_{\alpha}-K)\cup K) \leq m(E_{\alpha}-K) + m(K) \leq \epsilon + C < \infty$. So we found an absurd. – Manuel Bonanno Aug 26 '23 at 16:19
  • I am saying that $m(E)=\sup{m(K): K \text{ compact }, K\subset E}$ – Gio67 Aug 26 '23 at 22:49
  • https://math.stackexchange.com/questions/271848/inner-regularity-of-lebesgue-measurable-sets – Gio67 Aug 26 '23 at 22:56