Show that $\int|f(x)|dx=\int_0^\infty m(E_\alpha)d\alpha$

Let $f$ integrable and let $E_\alpha=\{x\mid |f(x)|>\alpha\}$ for $\alpha\geq 0$. Show that $$\int\left|f(x)\right|dx=\int_0^\infty m(E_\alpha)d\alpha$$ where $m$ is the Lebesgue measure.

By hypothesis, $f$ is measurable, therefore there exists a sequence of simple function such that $\varphi_n\nearrow |f|.$ Moreover, $E_\alpha^n\nearrow E^\alpha$. By the continuity of the measure, $m(E_\alpha^n)\to m(E_\alpha)$ for all $\alpha\geq 0$. By the convergence monotone theorem, $$\int_0^\infty m(E_\alpha)d\alpha=\lim_{n\to\infty }\int_0^\infty m(E_\alpha^n)\underset{(*)}{=}\lim_{n\to\infty }\int\varphi_n(x)dx=\int\left|f(x)\right|dx.$$

I don't understand the equality $(*)$, why $m(E^n_\alpha)=\varphi_n(x)$ ?

Let $\varphi_n = \sum_{k=1}^{N_n} a_k 1_{A_k}$, with positive $a_k$'s in increasing order and disjoint $A_k$'s, we have

$$m(E^n_\alpha) = \sum_{a_k > \alpha}m(A_k) = \sum_{k=j}^{N_n} m(A_k)$$ for $\alpha \in [a_{j-1},a_j), j = 1,2, \cdots, N_n$ with $a_{0} = 0$

So we have $$\int_0^\infty m(E^n_\alpha) d\alpha= \sum_{j=1}^{N_n}(a_j - a_{j-1})\sum_{k=j}^{N_n} m(A_k) = \sum_{j=1}^{N_n}a_j m(A_j) = \int \varphi_n(x)dx$$