Consider a random variable $X$ in $\mathcal{L}^1(\Omega,F,P)$ with distribution function $F(X,t)=P(X\leq t)$ for $t\in\mathbb{R}$.
Apparently, it is true that
$$\int_{-\infty}^\eta F(X,t)dt = \mathbb{E}(\eta - X)_{+}$$
where $(a)_+=\max(a,0)$. I am trying to prove this fact. These are my steps.
$$ \begin{align*} &\int_{-\infty}^\eta F(X,t)dt& = \int_{-\infty}^\eta \bigg[\int_{-\infty}^tP(d\beta)\bigg]dt \\ & & = \int_{-\infty}^t\bigg[\int_{-\infty}^\eta dt\bigg]P(d\beta)\\ & & = \color{red}{\cdots}\\ & & = \mathbb{E}(\eta - X)_{+} \end{align*} $$
In the second equality I use Fubini Theorem to switch the integrals, since the double integral is bounded. However, I am probably failing to capture something in the range of the integrals. Any hint on how to get from here to the expectation? It is easy to see that the outer integral is an expectation. But why the inner integral is $(\eta-X)_+$?
$$E(\eta-X)_+ = \int_0^\infty P((\eta-X)_+\geq t)\,dt = \lim_{T\to\infty}\int_0^T P(X \leq \eta-t)\,dt = \int_{-\infty}^\eta P(X\leq u)\,du $$