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The doubling map preserves normalized Lebesgue measure on the unit circle. Each point has exactly two preimages, $x/2$ and $(x+1)/2$ modulo one, and each inverse branch scales lengths by $1/2$.

For a Borel set $A\subset [0,1)$, the two inverse branches of the doubling map give the disjoint setsEach has half the Lebesgue measure of $A$, by translation and scaling invariance. Hence $\boxed{\mu (T^{-1}A)=\mu (A)}$. The piecewise affine map is Borel measurable, so it is a measure-preserving transformation. The printed one-dimensional “baker's map” is this doubling map.

Take the unit circle $X=\mathbb{R}/\mathbb{Z}$ with its Borel sigma-algebra and normalized Lebesgue measure, and letThis is the doubling map. Its two inverse branches halve length, so Lebesgue invariance of the doubling map gives $\mu (T^{-1}A)=\mu (A)$. Moreoverfor every point of the circle.

A map is Devaney chaos when:

Write a point as a binary expansionApart from the harmless choice of expansion at dyadic rationals, the binary shift representation of the doubling map is

Every open interval contains a binary cylinder specified by a finite initial word. Given cylinders $U$ and $V$, choose a binary sequence beginning with the word for $U$ and place the word for $V$ after it. A suitable iterate shifts the second word to the front, proving topological transitivity.

Given any cylinder, repeat its defining word forever. The resulting point is periodic and lies in that cylinder, so periodic points are dense.

Finally, given $x$ and any neighbourhood, choose $N$ so large that changing only digits after the first $N$ stays inside that neighbourhood. Choose the later tail so that after $N$ shifts it is either $0$ or $3/4$, whichever is farther from $F^N(x)$. The separation is at least $3/8$, so any smaller fixed constant, for example $\delta =1/4$, proves sensitivity. Hence the doubling map is chaotic in Devaney's sense.