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?

That is why you need the condition $$f$$ is integrable. If $$f$$ is integrable, you can deduce that $$a_j$$ is finite. 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 $$s_0. Then for any $$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.