Lebesgue Measure of Scalar Multiple

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Theorem

Let $\lambda^n$ be the $n$-dimensional Lebesgue measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\map \BB {\R^n}$.

Let $B \in \BB$.

Let $t \in \R_{>0}$.


Then:

$\map {\lambda^n} {t \cdot B} = t^n \map {\lambda^n} B$

where $t \cdot B$ is the set $\set {t \mathbf b: \mathbf b \in B}$.


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:

$\map \det {t \, \mathbf I_n} = t^n$


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

Thus, it follows that:

\(\ds \map {\lambda^n} {t \cdot B}\) \(=\) \(\ds \map {\lambda^n} {\map {\paren {t \, \mathbf I_n} } B}\)
\(\ds \) \(=\) \(\ds \map {\lambda^n} {\map {\paren {t^{-1} \, \mathbf I_n}^{-1} } B}\) Inverse of Group Inverse
\(\ds \) \(=\) \(\ds \map {\paren {\paren {t^{-1} \, \mathbf I_n}_* \lambda^n} } B\) Definition of Pushforward Measure
\(\ds \) \(=\) \(\ds \size {\map \det {\paren {t^{-1} \, \mathbf I_n}^{-1} } } \cdot \map {\lambda^n} B\) Pushforward of Lebesgue Measure under General Linear Group


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Now recall $\map \det {\paren {t^{-1} \, \mathbf I_n}^{-1} } = \map \det {t \, \mathbf I_n} = t^n$.

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

$\blacksquare$


Sources

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