# How is this book applying Fubini/Tonelli without assuming $\sigma$-finiteness?

I am reading about $L^p$ spaces on this google book and in proposition 1.1.4 (page 4) it writes $$p\int_0^\infty \alpha^{p-1} \int_X \chi_{\{x:|f(x)|>\alpha\}}d\mu(x)d\alpha = \int_X \int_0^{|f(x)|}p \alpha^{p-1}d\alpha d\mu(x)$$ stating "we used Fubini's theorem". Every version of Fubini/Tonelli I have ever seen assumes that both measures are $\sigma$-finite, even for nonnegative functions, and as far as I can tell no assumption on the $\sigma$-finiteness of $\mu$ was made here. What version of Fubini is this author applying?

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– user48738 Nov 9 '12 at 1:27
You are perfectly correct from the formal standpoint but if you think a bit, you'll realize that any $L^p(\mu)$ function is zero outside a set on which $\mu$ is $\sigma$-finite, so the rest of the space is just invisible in both integrals. – fedja Nov 9 '12 at 2:26
@fedja: "any Lp(μ) function is zero outside a set on which μ is σ-finite". Is $p \in (0, \infty]$ or $[1, \infty]$ or ...? Why is it? – Tim Feb 22 '13 at 20:10

Indeed, we have to work with $\sigma$-finite measures in order to apply Fubini's theorem. Although $\mu$ is not necessarily $\sigma$ finite, the set $\{x;|f(x)|>0\}$ can be written as a countable union of sets of finite measure. Namely, $\{x;|f(x)|>0\}=\bigcup_{n\geqslant 1}\{x,|f(x)|\geqslant n^{—1}\}$, and since $f$ is $\mu$-integrable, $\mu(\{x,|f(x)|\geqslant n^{—1}\})$ is finite. So we can work with $X':=\{x,|f(x)|\geqslant >0\}$ with trace $\sigma$-algebra and measure.