 Solution

Write $G=g_{s_1}\circ \cdots \circ g_{s_{n-1}}$. Because $g_{s_n}([0,1])\subseteq [0,1]$, taking the image under the same function composition givesThis uses preservation of set inclusion by every function, so decreasing inverse branches cause no difficulty. Since $f\circ g_L=f\circ g_R={\mathrm{id}}_{[0,1]}$, cancellation of the first inverse branch gives the equality of imagesThese compact intervals are itinerary cylinders: their symbols record successive positions under iteration of a map. In particular,Conversely these three memberships allow successive substitution of the appropriate inverse branches, and therefore characterize $I_{LLR}$ exactly.