How to integrate (and use Fubini Theorem) with variables on the boundary of the integrand? ($\int_0^\infty\int_0^x f(s)dsdx$) As the title goes, I am trying to calculate some norm and encountered integrals looking like 
$$\int_0^\infty\int_0^x f(s)dsdx$$
 where the integrand is a function 
$$F(x) = \int_0^x f(s)ds.$$
 I wonder if it's possible to use Fubini theorem here and do the integral with $x$ variable first. How does that affect the boundary of the inner integral?
 A: Assuming that Fubini's theorem can be applied then we have that
$$
\int_0^\infty \int_0^x f(s)\,\mathrm d s \,\mathrm d x=\int_{\Bbb R }\int_{\Bbb R }\mathbf{1}_{[0,\infty)}(x)\mathbf{1}_{[0,x]}(s) f(s)\,\mathrm d s \,\mathrm d x=\int_{\Bbb R ^2}\mathbf{1}_{A}(s,x)f(s)\,\mathrm d (s,x)
$$
for $A:=\{(s,x)\in \Bbb R ^2:0\leqslant s\leqslant x<\infty\}$, that is, a brief inspection show us that $\mathbf{1}_{A}(s,x)=\mathbf 1_{[0,\infty)}(x)\mathbf 1_{[0,x]}(s)$ because
$$
\mathbf{1}_{[0,\infty)}(x)\mathbf{1}_{[0,x]}(s)=1 \iff 0\leqslant x<\infty \,\land\, 0\leqslant s\leqslant x\iff 0\leqslant s\leqslant x<\infty 
$$
Before we was seeing $s$ as a variable who range depends of $x$, but now we want to see $x$ as the variable who range depends of $s$, therefore giving to $s$ full range we have that if $s\in[0,\infty )$ then to hold the condition that $0\leqslant s\leqslant x<\infty $ necessarily $x\in[s,\infty)$, so
$$
\int_{\Bbb R ^2}\mathbf{1}_{A}(s,x) f(s)\,\mathrm d (s,x)=\int_{\Bbb R^2}\mathbf{1}_{[0,\infty)}(s)\mathbf{1}_{[s,\infty)}(x) f(s) \,\mathrm d (s,x)
=\int_0^\infty \int_s^\infty f(s) \,\mathrm d x \,\mathrm d s
$$
