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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.

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Can you prove it for simple functions? –  Qiaochu Yuan Jan 15 '11 at 4:56
    
There are two typos. 1. the first instance of $\mu(g)$ should be $\mu(f)$. 2. In the integrals you can not have $E$ in both cases. –  AD. Jan 15 '11 at 6:04
    
My apologies - careless of me. Why can you not have E in both cases, sorry? If f has domain E (f: $E \to \mathbb{R}$, say) and $\tau$ is a map: $E \to E$, should the integrals not both be over E? Perhaps I am being slow. –  Giles H. Jan 15 '11 at 6:40
    
@Qiaochu, yes, I believe I can prove the first and second statement for simple functions (I originally wrote '$f \circ \tau$ is measurable' when I meant 'integrable', sorry for any confusion), but it is unclear to me why it follows necessarily that if for all simple functions f, $f \circ \tau$ is integrable, then for -all- functions f, $f \circ \tau$ is integrable. Is it perhaps because of the dominated convergence theorem, taking better and better simple approximations to f? –  Giles H. Jan 15 '11 at 6:45

1 Answer 1

up vote 1 down vote accepted

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.

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That's perfect, thankyou - yes, I'm very sorry, I appear to have written the question out a bit carelessly, it will be better next time. Thankyou so much for the help. –  Giles H. Jan 15 '11 at 22:28
    
@Giles H.: No problem ;-) Be sure to keep this argument in mind, it's of fundamental importance. A large number of basic results in measure theory are proved this way. –  t.b. Jan 15 '11 at 22:41

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