Problem:
Suppose $f$ is integrable on $\mathbb{R}^d$. For each $\alpha>0$, let $$E_\alpha=\{x:|f(x)|>\alpha\}.$$ Show that $$\int_{\mathbb{R}^d}|f(x)|\thinspace dx=\int_0^\infty m(E_\alpha)\thinspace d\alpha.$$ Hint: Consider the function $\chi_{(0,\infty)}(\alpha)\chi_{E_\alpha}(x)$, as a function of $(\alpha,x)$ in $\mathbb{R}\times\mathbb{R}^d$. Show the function is measurable and apply Tonelli's Theorem.
What I have tried so far:
I have shown the function $g(\alpha,x)=\chi_{(0,\infty)}(\alpha)\chi_{E_\alpha}(x)$ is measurable, and by Tonelli's Theorem, we have $$ \int_{\mathbb{R}^d}\int_\mathbb{R}g\thinspace d\alpha\thinspace dx=\int_\mathbb{R}\int_{\mathbb{R}^d}g\thinspace dx\thinspace d\alpha.$$ For any fixed $\alpha>0$, we have $$ \int_{\mathbb{R}^d}g\thinspace dx=\int_{\mathbb{R}^d}\chi_{E_\alpha}\thinspace dx=m(E_\alpha). $$ I think this will take care of the RHS of the desired result. However, I am unsure of how to proceed. For a fixed $x\in\mathbb{R}^d$, we have $$ \int_{\mathbb{R}^d}g\thinspace d\alpha=\int_0^\infty\chi_{E_\alpha}\thinspace d\alpha=\int_{F_x}1\thinspace d\alpha=m(F_x), $$ where $F_x=\{\alpha>0:|f(x)|>\alpha\}$. If for some reason $m(F_x)=|f(x)|$, we would obtain the LHS of the desired result. But I am not sure how to show this, or if it is even true. I feel like I am going down a weird rabbit hole, and would appreciate any hints as to how I should proceed.
$F_x$ is simply the interval $(0, |f(x)|)$ and the Lebesgue measure of this interval is its length, which is $|f(x)|$.