# Lebesgue integrability of the maximal function.

I'm practicing doing some questions on measure theory and I'm having great trouble with them. However, I've tried the following question and it seems quite easy hence I imagine I've probably made a mistake. I'm not a maths student (economics) so I'm finding this material tough.

I'm trying to show a set of equivalencies from the hardy littlewood maximal function. The question is:

If $\phi: \mathbb{R} \rightarrow [0,\infty)$ is a Borel measurable function, then $\phi\cdot\lambda^1$ denotes the measure defined by $$(\phi \cdot \lambda^1)(B)=\int_B \phi d\lambda^1$$ for all Borel sets $B \subseteq \mathbb{R}$.

If $\mu$ is a positive Borel measure on $\mathbb{R}$, then the maximal function $M(\mu):\mathbb{R}\rightarrow [0,\infty]$ of $\mu$ is defined by $$M(\mu)(x)=\sup_{r>0}\frac{\mu(B(x,r))}{\lambda^1(B(x,r))}$$

Let $f: \mathbb{R}\rightarrow \mathbb{R}$ be a Lebesgue integrable function. Show that the following are equivalent:

1. $M(|f|\cdot \lambda^1)=0$
2. $M(|f|\cdot \lambda^1)$ is Lebesgue integrable
3. $f=0$ Lebesgue-a.e.

$\textbf{Attempt at Solution}$

$(1)\Rightarrow(3)$: If $M(|f|\cdot \lambda^1)=0$ then it implies $(|f| \cdot \lambda^1)(B)=\int_B |f|\;d\lambda^1=0$. It follows that if $\int_B |f|\;d\lambda^1=0$ it must be that $f=0$ Lebesgue a.e.

$(1)\Rightarrow(2)$. If $M(|f|\cdot \lambda^1)=0$ then we know that $\int |M(|f|\cdot \lambda^1)|d\lambda^1<\infty$ and is therefore Lebesgue integrable.

Unfortunately I don't see how $(2)\Rightarrow(3)$ Am I doing this question in the right way or am I totally wrong? Thanks.

• Hint: If $\int_0^1|f|=c>0$ then $M|f|(t)\ge\dots$ for $t>1$. – David C. Ullrich Jan 10 '16 at 23:39

## 1 Answer

You seem to have limited the question to one dimension. Suppose that $A = \int_a^b |f|\,d\lambda^1 >0$ for some $a<b$. Then $$M(|f|\cdot\lambda^1)(x) = \sup_r {1\over 2r}\int_{x-r}^{x+r}|f|\,d\lambda^1\ge {A\over 2(x-a)}$$ for $x>b$, taking $r=x-a$. Because $\int_1^\infty dx/x$ diverges, we have $\int M(|f|\cdot\lambda^1)(x)\,d\lambda^1=\infty$. This shows that $M(|f|\cdot\lambda^1)$ integrable (i.e., having a finite integral) implies that $f=0$ a.e. in Lebesgue measure.