Translation in Euclidean Space is Measurable Mapping
Theorem
Let $\BB$ 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 $\BB \, / \, \BB$-measurable.
Proof
By Characterization of Euclidean Borel Sigma-Algebra, $\BB = \map \sigma {\JJ_{ho}^n}$.
Here, $\JJ_{ho}^n$ is the set of half-open $n$-rectangles, and $\sigma$ denotes generated $\sigma$-algebra.
Now, for any half-open $n$-rectangle $\horectr {\mathbf a} {\mathbf b}$, it is trivial that:
- $\map {\tau_{\mathbf x}^{-1} } {\horectr {\mathbf a} {\mathbf b} } = \horectr {\mathbf a + \mathbf x} {\mathbf b + \mathbf x}$
That is, the preimage of a half-open $n$-rectangle under $\tau_{\mathbf x}$ is again a half-open $n$-rectangle.
In particular, since $\JJ_{ho}^n \subseteq \map \sigma {\JJ_{ho}^n} = \BB$, Mapping Measurable iff Measurable on Generator applies.
Thus it follows that $\tau_{\mathbf x}$ is $\BB \, / \, \BB$-measurable.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 7$, $\S 7$: Problem $1$