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 $1$