measure of $X_B^+$ can only take a finite number of values. We write $\hat{N}(B)$ to show the dependence on $B$ explicitly.

\[thmxb+meas\] Suppose that $A=p/q$ and $\alpha=r/s$. For Lebesgue almost all $B$ we have $$\ell(X_B^+) = \frac{\hat{N}(B)}{qr}$$ where $\hat{N}(B)$ is an integer and $1\leq \hat{N}(B)\leq qrs$. If there is an interval on which $T$ has $L$ preimages, then $$\hat{N}(B)\leq qrs+1-L.$$

[[**Proof**]{} ]{}As preimages of $\pi^{-1}(J)$ are disjoint and all have length $\frac{1}{qr}$ we compute $$\ell(X_B^+) = \sum_{k=1}^{\hat{N}(B)} \frac{1}{qr} = \frac{\hat{N}(B)}{qr}.$$ Note that $\hat{N}(B)$ must take an integer value less than or equal to $qrs$, being the number of preimages $\pi^{-1}(\psi)$.

Since $T$ will be invertible on $X_B^+$, if there is an interval with more than one preimage, only one of these will be in $X_B^+$ and hence $\hat{N}(B)\leq qrs+1-L$ where $L$ is the number of preimages. [\
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We can now prove the main result in this section.

\[thmposmeas\] Suppose that $\alpha=r/s$ and $A=p/q$ are both rational, with $(r,s)=(p,q)=1$. Then the maximal invariant set $X^+$ has Hausdorff dimension $2$ and positive Lebesgue measure, more precisely, if there is an open set on which $f$ has $K$ preimages then $$\frac{1}{qrs}\leq \ell(X^+) \leq 1-\frac{K-1}{qrs}.$$

[[**Proof**]{} ]{}Note that under the hypotheses of the 