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.
This answer is just rewriting of the above comment by 'anthonyquas'.
In an Ergodic dynamical system $(X,\mathscr{{B}},\mu,T)$, $\mathbb{{E}_{\mu}}(f|T^{-1}\mathscr{B})=0$ may not imply $f=0$ a.e in $\mathscr{B}$. For example, consider $X=\{0,1\}^{\mathbb{N}}$ with the induced measure $\mu$ from the measure $\nu$ on $\{0,1\}$ defined by $\nu(0)=\nu(1)=\frac{1}{2}$, and $T:X\rightarrow X$ the left shift map defined by $T(x_{0},x_{1},x_{2},...)=(x_{1},x_{2},...)$. Then this system is Ergodic.
But, consider the continuous function $f:X\rightarrow\mathbb{R}$ defined by
$$ f(x_0,x_1,x_2, ...)= \begin{cases} 1 & \text{ if }x_0=0 \\ -1 & \text{ if }x_0=1 \end{cases} $$
Then for any $A\in\mathscr{B},T^{-1}A=\{0,1\}\times A$, and therefore
$$\int_{T^{-1}A}fd\mu=\int_{T^{-1}A\cap \{x_0=0\}}fd\mu + \int_{T^{-1}A\cap \{x_0=1\}}fd\mu=\frac{\mu (A)}{2}-\frac{\mu (A)}{2}=0.$$
So, in this case $\mathbb{E_{\mu}}(f|T^{-1}\mathscr{B})=0$ but $f\neq0$ in $\mathscr{B}$.