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 \begin{equation} (\phi \cdot \lambda^1)(B)=\int_B \phi d\lambda^1 \end{equation} 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 \begin{equation} M(\mu)(x)=\sup_{r>0}\frac{\mu(B(x,r))}{\lambda^1(B(x,r))} \end{equation}

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.

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

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.


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