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Given $U_T: L^2(X, \mu) \rightarrow L^2(X, \mu)$ , $f \mapsto f \circ T$ where $T : X \rightarrow X$ is measurable such that $\mu ( T^{-1}(A) ) = \mu(A) \forall A \in \Sigma$ I would like to show that $f = g$ $\mu$-almost-everywhere implies $U_T(f) = U_T(g)$ $\mu$-almost-everywhere.

My proof:

$$ \begin{align*} \| f\circ T - g \circ T \|_{L^2} &= \left(\int_X | f\circ T - g \circ T|^2 d \mu \right)^\frac{1}{2} \\&= \left(\int_{T(X)} | f- g |^2 d \mu \right)^\frac{1}{2} \\&= \left(\int_X | f- g |^2 d \mu \right)^\frac{1}{2} \\&= \| f-g\|_{L^2} \end{align*} $$

where the third equality comes from a change of variables.

Did I get this right? It feels a bit wobbly. Thanks for your help!

share|cite|improve this question
Well, that's right but somewhat overkill. Do it for simple functions first (which is straightforward from the assumptions); then extend by continuity to all of $L^2$ which you can do since the simple functions are dense. – t.b. Oct 17 '11 at 15:25
The third equality does not come from a change of variables. The second does, and you should integrate over $T^{-1}(X) = X$. The third equality is actually obvious. – André Caldas Oct 17 '11 at 17:15
I agree with what André says, I didn't read carefully enough, sorry about that. – t.b. Oct 17 '11 at 17:28
By the way, check the edits I did so you can improve your formatting. I used \begin{align*} and instead of ||, I used \|. And I also used \left and \right modifiers to your parenthesis. – André Caldas Oct 17 '11 at 18:49
@AndréCaldas: Thanks! – Rudy the Reindeer Oct 18 '11 at 6:42
up vote 1 down vote accepted

First of all, if a function $f: X\to \mathbb{R}$ is measurable so is $f \circ T: X \to \mathbb{R}$.

This is because by hypothesis $f^{-1}(B) \in \Sigma$ and $(f\circ T)^{-1}(B) = T^{-1}(f^{-1}(B))\in \Sigma$ for all Borel sets in $\mathbb{R}$ since $T:(X,\Sigma) \to (X,\Sigma)$ is measurable.

Now if $f: X\to \mathbb{R}$ is a function that is zero almost everywhere, there is a null set $N \in \Sigma$ such that $f(x) = 0$ for all $x \notin N$. But this means that $f(Tx) = 0$ for all $x \notin T^{-1}(N)$ and $\mu(T^{-1}(N)) = 0$ because of the assumption that $T$ preserves $\mu$. Therefore $f \circ T = 0$ almost everywhere.

Now if $f = g$ almost everywhere then $f-g = 0$ a.e., hence $(f-g)\circ T = f\circ T - g \circ T = 0$ a.e., hence $f \circ T = g\circ T$ almost everywhere.

That's what you wanted to show.

But now you want to go a step further and say that for $f \in L^2$ we have $\|f \circ T\|_2 = \|f\|_2$.

Your proof is flawed in that the very first expression already assumes that $f \circ T - g \circ T \in L^2$, and you don't know that yet!

What you do know after the first part of this answer is that $f = g$ a.e. implies that $f \circ T = g \circ T$ a.e. hence $|f \circ T|^2 = |g \circ T|^2$ a.e., and thus $f \circ T$ is square integrable if and only if $g \circ T$ is square integrable.

Now you want to prove that $\|f \circ T\|_2 = \|f\|_2$ for all $f \in L^2$, so here's an outline:

  1. If $f$ is simple, this is clear (why exactly?)
  2. If $f \geq 0$ then approximate $f$ monotonically with a sequence of simple functions $s_n$ and observe that $s_n \circ T \to f \circ T$ monotonically and apply the monotone convergence theorem and point 1.
  3. Split $f$ into real and imaginary parts and split real and imaginary parts into positive and negative parts and apply 2.
share|cite|improve this answer
Thank you! I was actually going to write it for simple functions as an answer in response to your comment. Regarding your comment: what does extend by continuity mean? Isn't it enough that simple functions are dense? – Rudy the Reindeer Oct 17 '11 at 18:26
Well, you need to extend $U_T$ from simple functions to all of $L^2$ and you can do that by appealing to continuity. But this does not a priori show that $U_T$ is given as composition $U_{T}f = f \circ T$ for all functions in $L^2$. However, the argument I sketched here gives that in addition, because $f \mapsto f \circ T$ it is a continuous extension of $U_T$ and the continuous extension is unique (by uniform continuity of linear operators). The fact I'm using is: – t.b. Oct 17 '11 at 18:31
If $X$ is a metric space, $Y$ is a complete metric space and $f: D \to Y$ is uniformly continuous with $D \subset X$ dense then there is a unique continuous extension of $f$ to all of $X$. – t.b. Oct 17 '11 at 18:32
I like it very much that you didn't just demonstrated that $f = g$ a.e. implies $f \circ T = g \circ T$ a.e. You also showed that $U_T$ is an isometry, because it seems that @Matt was mixing up both demonstrations. – André Caldas Oct 17 '11 at 18:42
Regarding your comment, the application $f \mapsto f \circ T$ is already well defined taking measurable functions "modulo a.e." to measurable functions "modulo a.e.". There is no need to appeal to simple functions. What remains to show, is that $U_T(L^2) \subset L^2$, and this is done by the second part of your proof. Notice that $\|\cdot\|_2$ is well defined for every measurable function, not just the ones in $L^2$. It's just that it might happen that it is not a finite value. – André Caldas Oct 17 '11 at 18:46

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