$f$ being measure preserving is to say that the measure of $f^{-1}(A)$ and the measure of $A$ are same for all measurable set $A$. I can see that $\int_X \lvert \phi\rvert \leq \int_X \lvert \phi\circ f\rvert$ since whatever cover $\{A_n\}_n$ we use to acquire integration on the left hand side, we can use its preimage $\{f^{-1}(A_n)\}_n$ to cover $X$ when we integrate $\lvert \phi\circ f\rvert$. I feel like two integrations should be equal so that $\phi\circ f$ is in $L^1$, but I don't know how to prove it. Do you think we need stronger condition on $f$ to ensure integrability of $\phi\circ f$? This question arises in the context of proving subadditive ergodic theorem, so I was thinking of $f$ being ergodic, but I'm not sure it's getting any better. Thanks for your help.
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typo in title, it should be $\phi\circ f$ – peek-a-boo Jun 24 '23 at 04:03
2 Answers
There’s a general change of variables formula:
If $(X,\mathfrak{M},\mu)$ is a measure space, $(Y,\mathfrak{N})$ a measurable space, and $f:X\to Y$ a measurable mapping, then for every measurable $\phi:Y\to [0,\infty]$, we have \begin{align} \int_X(\phi\circ f)\,d\mu&=\int_Y\phi\,d(f_*\mu), \end{align} where $f_*\mu$ is the push-forward measure defined by setting for each measurable set $E\in\mathfrak{N}$, $(f_*\mu)(E):=\mu(f^{-1}(E))$.
This is easily proved first when $\phi$ is an indicator of a measurable set (actually in this case it’s just the definition of the measure $f_*\mu$), then by linearity for non-negative simple functions, then by monotone convergence to all non-negative measurable $\phi$. Of course once we have it for all $\phi:Y\to [0,\infty]$, by taking the real and imaginary/positive/negative parts we can prove that $\phi\in L^1(Y,f_*\mu)$ if and only if $\phi\circ f\in L^1(X,\mu)$, and the two integrals are equal.
Now, in the special case that $X=Y$ and $\mathfrak{M}=\mathfrak{N}$, and $f$ is measure-preserving (i.e $f_*\mu=\mu$), you immediately get $\int_X(\phi\circ f)\,d\mu=\int_X\phi\,d\mu$.
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From the layer cake representation formula one has that $$ \int_{X} |\phi \circ f|(x) \mathsf{d} \mu (x)= \int_{0}^{\infty} \mu\left\{ \left( y: |\phi \circ f|(y)> \lambda \right\} \right) \mathsf{d} \lambda $$
Observe that $\left\{ y: |\phi \circ f|(y)> \lambda \right\}=f^{-1}\left( \left\{y: |\phi(y)| >\lambda \right\}\right) $ so by the measure preserving property of $ f $ there holds $$ \int_{0}^{\infty} \mu\left\{ \left( y: |\phi \circ f|(y)> \lambda \right\} \right) \mathsf{d} \lambda=\int_{0}^{\infty} \mu \left( \left\{y: |\phi(y)| >\lambda \right\} \right) \mathsf{d} \lambda = \int_{X} |\phi(x)| \mathsf{d} \mu(x) $$
