Translation in Euclidean Space is Measurable Mapping

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\mathcal B$ be the Borel $\sigma$-algebra on $\R^n$.

Let $\mathbf x \in \R^n$, and denote with $\tau_{\mathbf x}: \R^n \to \R^n$ translation by $\mathbf x$.


Then $\tau_{\mathbf x}$ is $\mathcal B \, / \, \mathcal B$-measurable.


Proof

By Characterization of Euclidean Borel Sigma-Algebra, $\mathcal B = \sigma \left({\mathcal{J}_{ho}^n}\right)$.

Here, $\mathcal{J}_{ho}^n$ is the set of half-open $n$-rectangles, and $\sigma$ denotes generated $\sigma$-algebra.


Now, for any half-open $n$-rectangle $\left[[{\mathbf a \,.\,.\, \mathbf b}\right))$, it is trivial that:

$\tau_{\mathbf x}^{-1} \left({\left[[{\mathbf a \,.\,.\, \mathbf b}\right))}\right) = \left[[{\mathbf a + \mathbf x \,.\,.\, \mathbf b + \mathbf x}\right))$

That is, the preimage of a half-open $n$-rectangle under $\tau_{\mathbf x}$ is again a half-open $n$-rectangle.


In particular, since $\mathcal{J}_{ho}^n \subseteq \sigma \left({\mathcal{J}_{ho}^n}\right) = \mathcal B$, Mapping Measurable iff Measurable on Generator applies.

Thus it follows that $\tau_{\mathbf x}$ is $\mathcal B \, / \, \mathcal B$-measurable.

$\blacksquare$


Sources