Euclidean Borel Sigma-Algebra Closed under Scalar Multiplication

Theorem
Let $\map \BB {\R^n}$ be the Borel $\sigma$-algebra on $\R^n$.

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

Then also $t \cdot B := \set {t \mathbf b: \mathbf b \in B} \in \BB$.

Proof
Define $f: \R^n \to \R^n$ by $\map f {\mathbf x} := \dfrac 1 t \mathbf x$.

Then for all $\mathbf x \in \R^n$, $\map {f^{-1} } {\mathbf x} = \set {t \mathbf x}$, where $f^{-1}$ denotes the preimage of $f$.

Thus $\ds t \cdot B = \bigcup_{\mathbf b \mathop \in B} \map {f^{-1} } {\mathbf b} = \map {f^{-1} } B$, where the last equality holds by definition of preimage.

It follows that the statement of the theorem comes down to showing that $f$ is $\BB \, / \, \BB$-measurable.

By Characterization of Euclidean Borel Sigma-Algebra the half-open $n$-rectangles $\JJ_{ho}^n$ generate $\BB$.

Applying Mapping Measurable iff Measurable on Generator, it suffices to demonstrate:


 * $\forall J \in \JJ_{ho}^n: \map {f^{-1} } J \in \BB$

Now for a half-open $n$-rectangle $J = \horectr {\mathbf a} {\mathbf b}$, it follows from the consideration on $f^{-1}$ above, that:


 * $\map {f^{-1} } J = \horectr {t \mathbf a} {t \mathbf b}$

which is again in $\JJ_{ho}^n$ and so, in particular, in $\BB$.

Thus $f$ is $\BB \, / \, \BB$-measurable, that is:


 * $\forall B \in \BB: t \cdot B = \map {f^{-1} } B \in \BB$

which was to be demonstrated.