This is a problem from Tao's blog.
Let ${f: {\bf R} \rightarrow {\bf R}}$ be an absolutely integrable function, and let ${f^*: {\bf R} \rightarrow {\bf R}}$ be the one-sided signed Hardy-Littlewood maximal function
$\displaystyle f^*(x) := \sup_{h>0} \frac{1}{h} \int_{[x,x+h]} f(t)\ dt$.
Show that $\forall \lambda > 0$,
$\displaystyle \lambda m( \{ f^*(x) > \lambda \} ) = \int_{x: f^*(x) > \lambda} f(x)\ dx$.
The hint suggest that we first try the case when ${f}$ has compact support, in which case one can apply the rising sun lemma to a sufficiently large interval containing the support of ${f}$. In applying the rising sun lemma to the function ${F: [a,b] \rightarrow {\bf R}}$ defined by $\displaystyle F(x) := \int_{[a,x]} f(t)\ dt - (x-a) \lambda$, I was trying to show that $a_n = a$ and $F(b_n) > F(a_n) = F(a)$ will lead to a contradiction when $\lambda > 0$, but was stuck. Any suggestion on how to proceed?