note that $$\begin{aligned}
T(N(\psi)) & \subseteq & \pi^{-1} \pi T(N(\psi)) = \pi^{-1} S \pi (N(\psi))\\
& = & \pi^{-1} S \pi (\pi^{-1} (\psi) \cap X_B^+) \\
& \subseteq  & \pi^{-1} S(\psi)) \cap X_B^+ \\
& =  & N(S(\psi)).\end{aligned}$$ For the other direction, suppose that $\theta\in N(S(\psi))$ and so $\pi(\theta)=S(\psi)$ and $\theta\in \cap_{n\geq 0} T^n(I)$. In particular, $\theta=T(\theta')$ for some $\theta'$. Now $\pi(\theta) = \pi T(\theta')= S \pi(\theta')$ and so $S(\psi)= S \pi(\theta')$. Invertibility of $S$ gives $\psi=\pi(\theta')$ and so $\theta'\in N(\psi)$. Hence $$T(N(\psi)) \supseteq  N(S(\psi))$$ and we have the result. [\
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The previous Lemma relies crucially on the fact that $S$ is invertible. The next result implies that the number of points in $N$ is a constant almost everywhere. Let $$\hat{N}=|N(\psi)|$$ be the cardinality of $N(\psi)$.

If $B$ is irrational then $\hat{N}$ is constant for a set of $\psi$ with full Lebesgue measure.

[[**Proof**]{} ]{}It is a standard result that $S_B$ is ergodic for Lebesgue measure if and only if $B$ is irrational. If we look at the set of $\psi$ that give a certain value of $N(\psi)$ this is invariant and therefore must have Lebesgue measure $0$ or $1$. [\
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One consequence of this is that the 