Lebesgue Measure is Invariant under Translations

Theorem
Let $\lambda^n$ be the $n$-dimensional Lebesgue measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\mathcal B \left({\R^n}\right)$.

Let $\mathbf x \in \R^n$.

Then $\lambda^n$ is translation-invariant; i.e., for all $B \in \mathcal B \left({\R^n}\right)$, have:


 * $\lambda^n \left({\mathbf x + B}\right) = \lambda^n \left({B}\right)$

where $\mathbf x + B$ is the set $\left\{{\mathbf x + \mathbf b: \mathbf b \in B}\right\}$.

Proof
Denote with $\tau_{\mathbf x}: \R^n \to \R^n$ the translation by $\mathbf x$.

From Translation in Euclidean Space is Measurable Mapping, $\tau_{\mathbf x}$ is $\mathcal B \left({\R^n}\right) \, / \, \mathcal B \left({\R^n}\right)$-measurable.

Consider the pushforward measure $\lambda^n_{\mathbf x} := \left({\tau_{\mathbf x}}\right)_* \lambda^n$ on $\mathcal B \left({\R^n}\right)$.

By Characterization of Euclidean Borel Sigma-Algebra, it follows that:


 * $\mathcal B \left({\R^n}\right) = \sigma \left({\mathcal{J}^n_{ho}}\right)$

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

Let us verify the four conditions for Uniqueness of Measures, applied to $\lambda^n$ and $\lambda^n_{\mathbf x}$.

Condition $(1)$ follows from Half-Open Rectangles Closed under Intersection.

Condition $(2)$ is achieved by the sequence of half-open $n$-rectangles given by:


 * $J_k := \left[{-k \,.\,.\, k}\right)^n$

For condition $(3)$, let $\lefthalf-open $n$-rectangle.

Since:


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

we have:

Finally, since:


 * $\displaystyle \lambda^n \left({J_k}\right) = \prod_{i \mathop = 1}^n \left({k - \left({-k}\right)}\right) = \left({2 k}\right)^n$

the last condition, $(4)$, is also satisfied.

Whence Uniqueness of Measures implies that:


 * $\lambda^n_{\mathbf x} = \lambda^n$

and since for all $B \in \mathcal B \left({\R^n}\right)$ we have:


 * $\mathbf x + B = \tau_{\mathbf x}^{-1} \left({B}\right)$

this precisely boils down to:


 * $\lambda^n \left({\mathbf x + B}\right) = \lambda^n \left({B}\right)$

Note
This theorem formalizes the physical intuition that the size of an object does not depend on its position.