Why $\int^{\infty}_{0}1_{|f(x)|>\alpha}d\alpha=|f(x)|$?

Question

$f$ is integrable on $\mathbb{R}^d$. For each $\alpha>0$, let $E_\alpha=\{x:|f(x)|>\alpha\}$. Show that $$\int_{\mathbb{R}^d}|f(x)|dx=\int^{\infty}_{0}m(E_\alpha)d\alpha$$

I saw a proof of the above that uses the fact that $$\int^{\infty}_{0}1_{|f(x)|>\alpha}d\alpha=|f(x)|$$

Why is this true? Intuitively it seems like we're "stacking" the values of the characteristic function until we get to the value of $f$.

Any help is much appreciated

Answer 1

$1_{\{|f(x)|>\alpha \}}$ is nothing but the charateristic function of the interval $(0,|f(x)|)$ so its integral equals the measure of this interval.