# Lebesgue Measure of Scalar Multiple

## Theorem

Let $\lambda^n$ be the $n$-dimensional Lebesgue measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\mathcal B \left({\R^n}\right)$.

Let $B \in \mathcal B$, and let $t \in \R_{>0}$.

Then $\lambda^n \left({t \cdot B}\right) = t^n \lambda^n \left({B}\right)$, where $t \cdot B$ is the set $\left\{{t \mathbf b: \mathbf b \in B}\right\}$.

## Proof

It follows from Rescaling is Linear Transformation that the mapping $\mathbf x \mapsto t \mathbf x$ is a linear transformation.

Denote $t \, \mathbf I_n$ for the matrix associated to this linear transformation by Linear Transformation as Matrix Product.

From Determinant of Rescaling Matrix, it follows that:

$\det \left({t \, \mathbf I_n}\right) = t^n$

From Inverse of Rescaling Matrix, $t \, \mathbf I_n$ is the inverse of $t^{-1} \mathbf I_n$.

Thus, it follows that:

 $\displaystyle \lambda^n \left({t \cdot B}\right)$ $=$ $\displaystyle \lambda^n \left({\left({t \, \mathbf I_n }\right) \left({B}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle \lambda^n \left({\left({t^{-1} \, \mathbf I_n }\right)^{-1} \left({B}\right)}\right)$ Inverse of Group Inverse $\displaystyle$ $=$ $\displaystyle \left({t^{-1} \, \mathbf I_n }\right)_* \lambda^n \left({B}\right)$ Definition of pushforward measure $\displaystyle$ $=$ $\displaystyle \left\vert{\det \, \left({ \left({t^{-1} \, \mathbf I_n }\right)^{-1} }\right) }\right\vert \cdot \lambda^n \left({B}\right)$ Pushforward of Lebesgue Measure under General Linear Group

Now recall $\det \, \left({\left({t^{-1} \, \mathbf I_n }\right)^{-1}}\right) = \det \, \left({t \, \mathbf I_n}\right) = t^n$.

Since $t > 0$, $\left\vert{t^n}\right\vert = t^n$, and the result follows.

$\blacksquare$