Solution

Ignore the zero-area boundary sets where a reciprocal or its digit is undefined, and put

a = ⌊ 1/ x ⌋

x = \frac{1}{a + x ^{'}}, y = \frac{1}{y ^{'}} - a, a = ⌊ \frac{1}{y ^{'}} ⌋ .

1/ (a + 1) < y^{'} < 1/ a

, covering the square almost everywhere. Thus the branch calculation proves global invariance without an extra sum over overlapping branches.

J = [x^{2} (a + y)^{2}]^{- 1}

. Since

1 + x^{'} y^{'} = \frac{a + y + 1/ x - a}{a + y} = \frac{1 + x y}{x ( a + y )},

the transformed density satisfies

P (x^{'}, y^{'}) J = \frac{1}{( l o g 2 ) ( 1 + x ^{'} y ^{'} ) ^{2}} \frac{1}{x ^{2} ( a + y ) ^{2}} = P (x, y) .

The change of variables formula therefore proves invariance. Normalization follows from

\int_{0}^{1} \frac{d y}{( 1 + x y ) ^{2}} = \frac{1}{1 + x}, \int_{0}^{1} \frac{d x}{( l o g 2 ) ( 1 + x )} = 1.

Projecting onto the first coordinate commutes with the Gauss continued-fraction map, so its invariant marginal is the Gauss measure:

p_{G} (x) = \frac{1}{( lo g 2 ) ( 1 + x )}, 0 < x < 1.

One may also check it directly. The inverse branches

h_{a} (x) = (a + x)^{- 1}

give

\sum_{a \geq 1} p_{G} (h_{a} (x)) ∣ h_{a}^{'} (x) ∣ = \frac{1}{l o g 2} \sum_{a \geq 1} \frac{1}{( a + x ) ( a + x + 1 )} = p_{G} (x),

where the last sum telescopes. Values assigned at the exceptional boundary points do not affect the absolutely continuous invariant measure.

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