is defined by $A=1/2$ and $\alpha=1$. In this case, we can approximate the maximal invariant set numerically as in Figure \[figa1\_2alp1\]. The map in this case is $$\label{eqeg0}
\begin{array}{l}
x'=\frac{3}{2}x+\frac{1}{2}y ~(\bmod 1)\\
y'=-\frac{1}{2}x+\frac{1}{2}y~(\bmod 1)
\end{array}$$ where $(x,y)\in X=[0,1]^2$. In this case, as noted in [@Zyc98], there appears to exist a symmetry between $X^+$ and its complement. In fact, we can use the results in Section \[secrational\] to get the result directly.

For the map (\[eqeg0\]) with $A=\frac{1}{2}$ and $\alpha=1$, we have $\ell(X^+)=\frac{1}{2}$.

[[**Proof**]{} ]{}Note that $r=s=p=1$ and $q=2$. Moreover, on $L_B$ the map (\[map1d\]) can be written $$\theta'=h(\theta)=\theta+\frac{B}{2} - \frac{1}{2}[B-\theta]~(\bmod 1)$$ and this map is two-to-one on an open set of $\theta$ for almost all $B$. Therefore, for almost all $B$, the upper and lower bounds of Theorem \[thmxb+meas\] agree and $$\ell(X_B^+)=\frac{1}{2}$$ for almost all $B$. This implies that $\ell(X^+)=\frac{1}{2}$. [\
]{}

Example II {#secegii}
----------

We now consider the case $A=\alpha=1/2$; this shows that $\ell(X^+)$ may be irrational for a rational map. In this case the map is $$\label{eqeg1}
\begin{array}{l}
x'=\frac{3}{2}x+y ~(\bmod 1)\\
y'=-\frac{1}{4}x+\frac{1}{2}y~(\bmod 1)
\end{array}$$ On the invariant lines $L_B$ with $B\in[\frac{1}{2},1)$ (\[eqeg1\]) reduces to (\[map1d\]) with $\theta\in[0,2)$ to $$\theta' = T_B(\theta)=\left(\theta+B-\frac{1}{2}[\theta]-
[B-2\theta] \right) ~ (\bmod 2).
\label{pmap}$$

