Measure theoretic proof on composition of invariant functions and integrability I'm looking for a proof that if f is $\mu$-integrable ($\mu(|f|) < \infty$, where $\mu(f)$=sup{$\mu(g):g \leq f,\ g  \text{ simple}$}), and $\tau$ is measure preserving ($\tau^{-1}(A)$ measurable for every measurable A), then $f \circ \tau$ is integrable and $\int f d\mu = \int f \circ \tau \, d\mu$. I have tried proving it myself (I need a proof but I don't need to have proved it myself) but I'm not sure where to start. Could anyone either direct me to a proof (online if possible though if it's a book that's not a disaster) or help me with such a proof? I can't seem to get going on it, so the more help you can give the better!
Thanks very much, Giles. 
 A: There are two issues in the formulation of your question:


*

*The definition
\[
\mu(f) = \sup\{\mu(g)\,:\,\text{$g \leq f$ simple}\}
\]
only works for $f \geq 0$. For general $f$ you have to write $f = f_+ - f_{-}$ with $f_+ \geq 0$ and $f_{-} \geq 0$ and use that by definition $\mu(f) = \mu(f_{+}) - \mu(f_{-})$.

*You forgot the condition $\mu(A) = \mu(\tau^{-1}A)$ in the definition of measure-preserving.
As Qiaochu pointed out, the case of simple functions implies the general case. I'll treat the case $f \geq 0$ only.
Note that for the characteristic function $g = [A]$ of a measurable set $A$ we have $g \circ \tau = [\tau^{-1} A]$ and $\mu(g) = \mu(g \circ \tau)$ by hypothesis. By linearity of the integral $\mu(g) = \mu(g \circ \tau)$ for simple $g$. Now if $f \geq 0$ we may choose a sequence $g_{n}$ of simple functions such that $g_{n} \to f$ pointwise (a.e.) and monotone. By using the monotone convergence theorem twice and using the fact that $\tau$ is measure-preserving in the second equality we have
\[
\mu(f) = \lim_{n \to \infty} \mu(g_{n}) = \lim_{n \to \infty} \mu(g_{n} \circ \tau) = \mu(f \circ \tau)
\]
as desired.
