*I am trying to figure out how to answer this question in full.
Recall, that if µ is a complex Borel measure on R, then the maximal function $M(µ) : \Re \to [0, 1]$ of µ is defined by $M(µ)(x) = \sup_{r>0}\frac{|µ|(B(x, r))} {\lambda^1(B(x, r))} $, where $\lambda$ denotes a Lebesgue measure in R.
Also, if $\phi : R \to [0, 1)$ is a Borel measurable function, then $\phi · \lambda$ denotes the measure defined by $(\phi · \lambda)(B) = \int_B\phi d \lambda$ for all Borel sets $B\subset \Re$
Let $f : R \to R$ be a Lebesgue integrable function. Show that the following statements are equivalent. (a) $M(|f| · \lambda) = 0.$
(b) $M(|f| · \lambda)$ is Lebesgue integrable
(c) $f=0$ Lebesgue a.e.*
Solution
I have been able to show that $a \iff c$ but I am struggling on how to start showing that $a \iff b$ and $b \iff c$.
My answer thus far: We first note that $M(\mu)(x)=\sup_{r>0}\frac{|\mu|(B(x,r))}{\lambda(B(x,r))}=\sup_{r>0}\frac{1}{\lambda(B(x,r))}\int_{B(x,r)}|f|d\lambda$
Then we suppose that c holds such that, $M(\mu)(x)=\sup_{r>0}\frac{|\mu|(B(x,r))}{\lambda(B(x,r))}=\sup_{r>0}\frac{1}{\lambda(B(x,r))}\int_{B(x,r)}|f|d\lambda=0$, then it is clear that a is also true.
Then for the converse we assume a is true yielding, $0=\int_{B(x,r)}|f|d\lambda$, so we are able to deduce that f is Lebesgue measurable a.e. as it is non-negative.
Once we establish $b \Rightarrow c$, then the rests follow directly. Assume to the contrary that $|f|>0$ on a set of positive measure. Then there exists a ball $B\subset\mathbb{R}^n$ such that $$ \int_B |f(x)|dx=c >0. $$ Without loss of generality, assume that $B= B(0,1) =\{x\in\mathbb{R}^n\;|\;\|x\|<1\}$. Then, for $y$ with $\|y\|=r$, we have $$\begin{eqnarray} M(|f|\cdot\lambda)(y) &\geq& \frac{1}{|B(y,r+1)|}\int_{B(y,r+1) }|f(x)|dx\\&\geq& \frac{1}{|B(y,r+1)|}\int_B |f(x)|dx\\ &\geq& c'\cdot r^{-n}=c'\|y\|^{-n}, \end{eqnarray}$$ for some constant $c'>0$ and for all sufficiently large $r>0$. Conclude from this fact that $M(|f|\cdot\lambda)\notin L^1(\mathbb{R}^n)$.