Lebesgue Measure of Scalar Multiple
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
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 5$: Problem $8$
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 7$: Problem $7$