Pushforward of Lebesgue Measure under General Linear Group

Theorem
Let $M \in \operatorname{GL} \left({n, \R}\right)$ be an invertible matrix.

Let $\lambda^n$ be $n$-dimensional Lebesgue measure.

Then the pushforward measure $M_* \lambda^n$ satisfies:


 * $M_* \lambda^n = \left\vert{\det M^{-1}}\right\vert \cdot \lambda^n$

Proof
From Linear Transformation on Euclidean Space is Continuous, $M^{-1}$ is a continuous mapping.

Thus from Continuous Mapping is Measurable, it is measurable, and so $M_* \lambda^n$ is defined.

Now let $B \in \mathcal B \left({\R^n}\right)$ be a Borel measurable set, and let $\mathbf x \in \R^n$.

Then:

Thus $M_* \lambda^n$ is a translation-invariant measure.

From Translation-Invariant Measure on Euclidean Space is Multiple of Lebesgue Measure, it follows that:


 * $M_* \lambda^n = M_* \lambda^n \left({\left({0 \,.\,.\, 1}\right)^n}\right) \cdot \lambda^n$

Lastly, using Determinant as Volume of Parallelotope it follows that:


 * $M_* \lambda^n \left({\left({0 \,.\,.\, 1}\right)^n}\right) = \lambda^n \left({M^{-1} \left({\left({0 \,.\,.\, 1}\right)^n}\right)}\right) = \left\vert{\det M^{-1}}\right\vert$

Hence the result.