Euclidean Borel Sigma-Algebra Closed under Scalar Multiplication

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Theorem

Let $\mathcal B \left({\R^n}\right)$ be the Borel $\sigma$-algebra on $\R^n$.

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


Then also $t \cdot B := \left\{{t \mathbf b: \mathbf b \in B}\right\} \in \mathcal B$.


Proof

Define $f: \R^n \to \R^n$ by $f \left({\mathbf x}\right) := \dfrac 1 t \mathbf x$.

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

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


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

By Characterization of Euclidean Borel Sigma-Algebra the half-open $n$-rectangles $\mathcal{J}_{ho}^n$ generate $\mathcal B$.

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

$\forall J \in \mathcal{J}_{ho}^n: f^{-1} \left({J}\right) \in \mathcal B$

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

$f^{-1} \left({J}\right) = \left[[{t \mathbf a \,.\,.\, t \mathbf b}\right))$

which is again in $\mathcal{J}_{ho}^n$ and so, in particular, in $\mathcal B$.


Thus $f$ is $\mathcal B \, / \, \mathcal B$-measurable, i.e.:

$\forall B \in \mathcal B: t \cdot B = f^{-1} \left({B}\right) \in \mathcal B$

which was to be demonstrated.

$\blacksquare$


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