# Pushforward of Lebesgue Measure under General Linear Group

## Theorem

Let $M \in \GL {n, \R}$ be an invertible matrix.

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

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

$M_* \lambda^n = \size {\det M^{-1} } \cdot \lambda^n$

## 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 $B \in \map {\mathcal B} {\R^n}$ be a Borel measurable set, and let $\mathbf x \in \R^n$.

Then:

 $\displaystyle \map {M_* \lambda^n} {\mathbf x + B}$ $=$ $\displaystyle \map {\lambda^n} {\map {M^{-1} } {\mathbf x + B} }$ Definition of Pushforward Measure $\displaystyle$ $=$ $\displaystyle \map {\lambda^n} {\map {M^{-1} } {\mathbf x} + \map {M^{-1} } B}$ $M^{-1}$ is linear $\displaystyle$ $=$ $\displaystyle \map {\lambda^n} {\map {M^{-1} } B}$ Lebesgue Measure is Translation-Invariant $\displaystyle$ $=$ $\displaystyle \map {M_* \lambda^n} B$ Definition of Pushforward Measure

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

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

$M_* \lambda^n = \map {M_* \lambda^n} {\openint 0 1^n} \cdot \lambda^n$

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

$\map {M_* \lambda^n} {\openint 0 1^n} = \map {\lambda^n} {\map {M^{-1} } {\openint 0 1^n} } = \size {\det M^{-1} }$

Hence the result.

$\blacksquare$