Suppose that $f$ is integrable on $\mathbb{R}^d$. For each $\alpha >0$, let $E_{\alpha}= \{ x : |f(x)| > \alpha\}$
Prove that $$\int_{\mathbb{R}^d} |f(x)| dx = \int_{0}^{\infty} m(E_{\alpha}) d\alpha$$.
$\textbf{My Attempt}:$ I think this is an application of Fubini's thoerem.
Also, $$ \int_{0}^{\infty} \int_{\mathbb{R}} \chi(x)_{E_{\alpha}} dx d\alpha = \int_{0}^{\infty} m(E_{\alpha})d \alpha$$
So I want to use something along these lines but I am stuck
You are very close. $m(E_\alpha) = \displaystyle \int_{\mathbb R^d} \chi_{E_\alpha}(x) \, dx$ so by the Tonelli theorem $$m(E_\alpha) = \int_{\mathbb R^d} \int_0^\infty \chi_{E_\alpha}(x) \, d\alpha dx.$$Note that $\chi_{E_\alpha}(x) = 1$ if and only if $\alpha < |f(x)|$ and is zero otherwise, so that $$ \chi_{E_\alpha}(x) = \chi_{[0,|f(x)|)}(\alpha)$$ and $$ \int_0^\infty \chi_{E_\alpha}(x) \, d\alpha = \int_0^\infty \chi_{[0,|f(x)|)}(\alpha) \, d\alpha = \int_0^{|f(x)|} \, d\alpha = |f(x)|.$$