Prove that $\int_X |f| d\mu = \int_{0}^{\infty} \lambda(\alpha)d\alpha$

Question

Here, I have old question on distribution function.

Let $(X,\mathcal{M},\mu)$ be a measure space and let $\lambda(\alpha) = \mu(\left\{x\in X : |f(x)|>\alpha \right\})$.

Firstly, prove that, for $f\in L^1(X)$,

$\int_X |f| d\mu = \int_X \left(\int_{0}^{|f|} d\alpha \right) d\mu$.

After that, by using the above, prove that

$\int_X |f| d\mu = \int_{0}^{\infty} \lambda(\alpha)d\alpha$.

Answer 1

Hint:

$\int_{0}^{\infty}\lambda\left(\alpha\right)d\alpha=\int_{0}^{\infty}\int_{X}1_{\left(\alpha,\infty\right)}\left(\left|f\right|\right)d\mu d\alpha$

Now switch the order of integration.