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?

  • 1
    $\begingroup$ I have edited my answer giving more details. In general its easy to see how to change the order of integration working with indicator functions because product of indicators functions are easy to analyze and see how to write them in a different way $\endgroup$
    – Masacroso
    Feb 2, 2020 at 18:42
  • $\begingroup$ related math.stackexchange.com/q/1489737/532409 $\endgroup$
    – Quillo
    Apr 29 at 17:34

1 Answer 1


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 $$ \int_{\Bbb R ^2}\mathbf{1}_{A}(s,x) f(s)\,\mathrm d (s,x)=\int_{\Bbb R^2}\mathbf{1}_{[0,\infty)}(s)\mathbf{1}_{[s,\infty)}(x) f(s) \,\mathrm d (s,x) =\int_0^\infty \int_s^\infty f(s) \,\mathrm d x \,\mathrm d s $$


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