Prove that if $E_{\alpha}^+=$ {$x \in R :f_+^*(x) \gt \alpha$} ,then $m(E_{\alpha}^+)=\frac{1}{\alpha} \int_{E_{\alpha}^+} \vert f(y) \vert \, dy$ 
We define the "one-side" maximal function $$f_+^*(x)=\sup_{0 \lt
 h}\frac{1}{h} \int_x^{x+h} \vert f(y) \vert \, dy$$ Prove that if $f$
   is integrable and $E_{\alpha}^+=$ {$x \in R :f_+^*(x) \gt \alpha$}
   ,then $$m(E_{\alpha}^+)=\frac{1}{\alpha} \int_{E_{\alpha}^+} \vert
 f(y) \vert \, dy \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$ [Hint: Apply Rising Sun lemma to $F(x)= \int_0^x
 \vert f(y) \vert \,dy
 -\alpha x$].  

If $x \in E_\alpha^+$, it implies that for each x there is $\, h_x$, so that $F(x+h_x) \gt F(x)\,$  and By the Rising Sun lemma,$$E_\alpha^+ =\cup_{j=1}^\infty (a_j,b_j)$$   And we have $\int_{a_j}^{b_j}
 \vert f(y) \vert \,dy = \alpha(a_j-b_j)\,$ if $\,(a_j,b_j)$ is a finite interval.If both  intervals are finite interval,sum these equation we can get the desired conclusion.
However, if $a_j=-\infty$ or $b_j=\infty$,we can not get $\int_{a_j}^{b_j}
 \vert f(y) \vert \,dy = \alpha(a_j-b_j)\,$ , How to deal with this condition?Another question,can the equation (1) hold when we do not assume $f$ is integrable?
 A: That is why you need the condition $f$ is integrable. If $f$ is integrable, you can deduce that $a_j$ is not infity. Call your $f^*$ as $M_Rf$.
For any $x\in (a_i,b_i)$, define
$$
N_x=\{s\in (x, b_i]: \int_{x}^s |f(y)|dy > \alpha (s- x)\}.
$$
Claim: $N_x$ is nonempty and $\sup_s N_x=b_i$ for any $x\in (a_i,b_i)$.
For any $x\in (a_i,b_i)$,
$$
M_Rf(x)=\sup_{r}\cfrac{1}{r}\int_x^{x+r}|f(y)|dy>\alpha,
$$
Thus there exist $r_0>0$ such that $\frac{1}{r_0}\int_x^{x+r_0}|f|dy>\alpha$, set $\alpha=r_0$. Hence $N_x$ is nonempty. Since $s\le b_i$, $s_0=\sup_s N_x$ exists. We now assume that $s_0<b_i$. Then for any $t\in (s_0,b_i]$, we have
$$
\int_{s_0}^t|f(y)|dy\le \alpha(t-s_0).
$$
Otherwise $\int_x^t |f(y)|dy> \alpha(s_0-x)+\alpha(t-s_0)=\alpha(t-x)$. But we know that
$M_R(s_0)>\alpha$. Thus there exists $r_{s_0}>b_i$ such that
$$
\int_{s_0}^{r_{s_0}}|f(y)|dy > \alpha(r_{s_0}-s_0).
$$
Thus $\int_{b_i}^{r_{s_0}}|f(y)|dy>\alpha(r_{s_0}-b_i)$, which implies that $M_R(x)>\alpha$ when $x\in [b_i,r_{s_0})$. This disagrees the definition of $b_i$.
The claim is completed and then we know
$$
\int_{a_i}^{b_i}|f(y)|dy \ge \alpha(b_i-a_i).
$$
This also tells us that $a_j\not = -\infty$ since $f$ is integrable.
