swapping integration order in renewal theory

Question

So I was studying renewal theory for markov chains and this appeared in one of the slides. F(x) is a CDF of some not specified random variable. $$ x \vee s$$ means a maximum between x and s. My only doubt probably emerges from lack of some fundamental basis. I simply don't understad how the integration order was changed, especially about the bound from 0 to u. If someone could show me more detailed steps, explanation how it was done (logic behind it) it would be great. I am aware that question is probably really basic. Thanks for any help. I think I generally know how to change order of integration but in this case I am lost. Relevant below: $$ \begin{array}{c} \frac{1}{\mu} \int_{0}^{\infty}(1-F(x \vee s)) d s= \\ \int_{0}^{\infty} \int_{x \vee s}^{\infty} d F(u) d s=\int_{x}^{\infty} \int_{0}^{u} d s d F(u)=\int_{x}^{\infty} u d F(u) \end{array}$$

Answer 1

This is a straight calculus question (no probability). You are integrating over the shaded area. Either you: fix $u$, and for each fixed $x \leq u < \infty$, you go between the blue curves, or you fix $s$, and if $s \leq x$ you go from $x$ to $\infty$, and if $s > x$ you go from $s$ to $\infty$.

$$\int_x^\infty \int_{\color{blue}{0}}^{\color{blue}{u}} \dots ds F(du) = \int_0^\infty \int_{\color{red}{x\vee s}}^\infty \dots F(du) ds.$$