# A continuous area-preserving mapping is an isometry?

Suppose that $f\colon\mathbb R^2\to\mathbb R^2$ is a continuous map which preserves area in the Euclidean sense. Can we say that $f$ is an isometry?

Note. We donot assume that $f$ is differentiable.

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Try $f$ linear with determinant $1$. – Gerry Myerson Jan 24 '13 at 11:44
@Avatar, the MO question is interesting, but not directly relevant, is it? – Gerry Myerson Jan 24 '13 at 11:55
@GerryMyerson: i understood your point,sorry; i am deleting my comment. – Aang Jan 24 '13 at 12:11

consider the matrix $$A = \left( \begin{array}{ccc} 2 &0 \\ 0 & 1/2 \\ \end{array} \right)$$ and the continuous function $y=Ax$.
The area of any set remains the same under this linear transformation, but $|A \hat{x}| = 2 > 1.$