probability - Expected value of a continuous random variable (understanding of the proof)

I need a little help to understand the steps of this proof. (I'm currently studying double integrals) but also i don't understand the first equalities. Thank you in advance.

Let $X$ be an absolutely continous variable with density function $f_x$. Suppose that $g$ is a nonnegative function then.
$$E(g(X))=\int_{0}^{\infty}P(g(X))>y)dy-\int_{0}^{\infty}P(g(X)\leqslant -y)dy$$
$$=\int_{0}^{\infty}P(g(X))>y)dy=\int_{0}^{\infty} (\int_{B} f_x(x) dx)dy$$

where B:= {x:g(x)>y}. Therefore.
$$E(g(X))=\int_0^\infty \int_0^g(x) f_x dydx= \int_0^\infty g(x)f_x(x)dx.$$

Answer

The first equality can be skipped if you just use the tail sum formula for nonnegative random variables: $E[Y] = \int_0^\infty P(Y > y) \mathop{dy}$ if $Y$ is continuous and nonnegative. Applying this to $Y=g(X)$ immediately yields $E[g(X)] = \int_0^\infty P(g(X)) > y) \mathop{dy}$.

[The first equality is a generalization, and can be proven by writing $Y=Y \cdot 1_{Y \ge 0} - (-Y) \cdot 1_{Y < 0}$ and applying the tail sum probability to each term both of which are nonnegative.]

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The next part is simply the definition of $B$. $$P(g(X)>y) = \int_B f_X(x) \mathop{dx}$$

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Finally, switch the order of the two integrals.
I think you forgot to mention that $X$ is also nonnegative?
The region you are integrating over is $\{y \ge 0\} \cap \{x \ge 0 :g(x) > y\}$, which can be rewritten as $\{x \ge 0\} \cap \{y : 0 \le y < g(x)\}$.

$$\int_0^\infty \int_B f_X(x) \mathop{dx} \mathop{dy} = \int_0^\infty \int_0^{g(x)} f_X(x) \mathop{dy} \mathop{dx}.$$