 Solution

Ignore the zero-area boundary sets where a reciprocal or its digit is undefined, and put $a=\lfloor 1/x\rfloor$. On this digit branch the natural extension of the continued-fraction Gauss map has inverseThe images are the disjoint horizontal strips $1/(a+1)<y^{\prime }<1/a$, covering the square almost everywhere. Thus the branch calculation proves global invariance without an extra sum over overlapping branches.

Its Jacobian determinant has magnitude $J=[x^2(a+y{)}^2{]}^{-1}$. Sincethe transformed density satisfiesThe change of variables formula therefore proves invariance. Normalization follows fromProjecting onto the first coordinate commutes with the Gauss continued-fraction map, so its invariant marginal is the Gauss measure:One may also check it directly. The inverse branches $h_a(x)=(a+x{)}^{-1}$ givewhere the last sum telescopes. Values assigned at the exceptional boundary points do not affect the absolutely continuous invariant measure.