I am following this document http://www.jstor.org/stable/2037431?seq=4
Shouldn't it be necessary to check that the chacon transform is ergodic? The theorem I'm familiar is this:
Let $T$ be a bijective measure preserving transform on a probability space. Then $T$ is weak mixing if and only if the induced operator $U$ on $L^2$ (Complex $L^2$) has point spectrum consisting exactly of the point $1$ with each eigenvector for $1$ being the constant class.
I see that they have verified that the only eigenvalue is $1$, and I think I understand it although some details need to be hammered out in my head. But where have they checked that $1$ only has constant eigenvectors? In other words, where have they checked that this $T$ is ergodic? If they haven't checked this, can someone prove it please?
I should say that I always use $T$ to denote the measure preserving transform in question, even though they have used $\sigma$ for this.
As an extra curiosity, I wonder if the theorem I know characterizing weak mixing in terms of complex $L^2$ can also be generalized to situations where $T$ is not a bijective measure preserving transform.
Edit: Actually I no longer think I understand the details surrounding the section on the 4th page where the document constructs an $\alpha$ such that on 3/4 of some interval, one has an approximation. I don't get why such an $\alpha$ must exist, so if someone could pick up the argument from there that would be helpful too. Answers to any subset of my 3 questions are appreciated.
Edit: I see, I only get free access because I'm at a university. Okay, here's a self-contained subquestion that might give me enough of a push to get going on two of the questions. I'd still have the question about generalizing
"Let $T$ be a bijective measure preserving transform on a probability space. Then $T$ is weak mixing if and only if the induced operator $U$ on $L^2$ (Complex $L^2$) has point spectrum consisting exactly of the point $1$ with each eigenvector for $1$ being the constant class."
to nonbijective mpts, but that's self contained.
How do I show this: Let $f \in L^2[0, 1]$ with the Borel $\sigma$-algebra and lebesgue measure. Then there exists $\alpha\neq0$ so that for all $\epsilon>0$ there exist $k, n \geq 0$ with $k<3^{(n+1)}$ such that there exists a borel $J \subset [k/3^{(n+1)}, (k+1)/3^{(n+1)})$ with $|J| \leq min(1, \epsilon)/(8*3^{(n+1)})$ and on $[k/3^{(n+1)}, (k+1)/3^{(n+1)}) - J$ we have $|f-\alpha|<\epsilon$?