A derivation is given in a textbook. Here, $F$ is a probabilistic distribution function, which means that: $$\int_{-\infty}^{\infty} F(t) dt = 1$$
At one step, they change the order of integration:
\begin{align} H(x) &= \int_{-\infty}^x \int_s^x dt \: dF(s) \\ &= \int_{-\infty}^x \int_{-\infty}^t dF(s) \: dt \end{align}
My question is, how did they come up with the bounds in the second line?
Given a function $ f : \mathbb{R}^{2}\rightarrow\mathbb{R} $, such that $ \left(\forall s\in\mathbb{R}\right),\ f_{s}:t\mapsto f\left(t,s\right) $ is continious and integrable in $ \mathbb{R} $, and that $ g_{s}:x\mapsto\int\limits_{x}^{+\infty}{f\left(t,s\right)\mathrm{d}t} $ is also continious and integrable in $ \mathbb{R},\ \left(\forall s\in\mathbb{R}\right)\cdot $
Let $ x $ be a real.
For $ s \leq x $, denoting $ \left(\forall t\in\left]-\infty,x\right]\right),\ g_{s}\left(t\right)=\left\lbrace\begin{aligned}f\left(t,s\right),\ \ \ \textrm{If }t\geq s\\ 0,\ \ \ \textrm{If }t\leq s\end{aligned}\right. $, we have :
$$ \int_{-\infty}^{x}{\int_{s}^{x}{f\left(t,s\right)\mathrm{d}t}\,\mathrm{d}s}=\int_{-\infty}^{x}{\int_{-\infty}^{x}{g_{s}\left(t\right)\mathrm{d}t}\,\mathrm{d}s} $$
Using Fubini's theorem we get : \begin{aligned} \int_{-\infty}^{x}{\int_{s}^{x}{f\left(t,s\right)\mathrm{d}t}\,\mathrm{d}s}&=\int_{-\infty}^{x}{\int_{-\infty}^{x}{g_{s}\left(t\right)\mathrm{d}s}\,\mathrm{d}t}\\ &=\int_{-\infty}^{x}{\int_{-\infty}^{t}{f\left(t,s\right)\mathrm{d}s}\,\mathrm{d}t} \end{aligned}