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Let us consider a measure preserving dynamical system $(X,\mathscr{B},\mu,T)$, where $X$ is a compact metric space, $\mathscr{B}$ is a $\sigma$- algebra which contains the Borel $\sigma$-algebra on $X$ and $T:X\rightarrow X$ is continuous.

Then for some fixed $\:f:X\rightarrow \mathbb{R}$ continuous, can we say that $$\mathbb E_\mu(f|T^{-1}\mathcal B)=0 \implies f = 0$$

[ This question came up in my mind when studying the proof of Proposition 3.4 from 'Recurrence in Ergodic Theory and Combinatorial Number Theory' by Harry Furstenberg. ]

Thanks in advance for any help.

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    $\begingroup$ No: let $X=\{0,1\}^{\mathbb Z_{\ge 0}}$ and $\mu$ be the coin-tossing meaure. Let $\mathcal B$ be the Borel $\sigma$-algebra and let $f(x)=1$ if $x_0=1$; and $-1$ id $x_0=1$. Then $\mathbb E_\mu(f|T^{-1}\mathcal B)=0$. Proof: let $A$ be a subset of $X$. Then $T^{-1}A=0A\cup 1A$, where $iA=T^{-1}A\cap \{x\colon x_0=i\}$. Now $\mu(iA)=\frac 12\mu(A)$ (you can check this for cylinder sets and then it follows for all sets). Now $\int_{T^{-1}A}f\,d\mu=\int_{0A}f\,d\mu+\int_{1A}f\,d\mu=\mu(0A)-\mu(1A)=(\frac 12-\frac 12)\mu(A)=0$. $\endgroup$ – anthonyquas May 29 '18 at 16:48
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    $\begingroup$ Since $\int_Z f\,d\mu=0$ for any $Z\in T^{-1}\mathcal B$, you deduce $\mathbb E(f|T^{-1}\mathcal B)=0$. $\endgroup$ – anthonyquas May 29 '18 at 16:54
  • $\begingroup$ @anthonyquas Thank you. $\endgroup$ – Surajit May 29 '18 at 20:43

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