Lebesgue Measure 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 $\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \in \mathcal{J}^n_{ho}$ be a half-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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \lambda^n_{\mathbf x} \left({\left[\left[{\mathbf a \,.\,.\, \mathbf b}\right)\right)}\right)\)	\(=\)	\(\displaystyle \lambda^n \left({\left[\left[{\mathbf {a + x} \,.\,.\, \mathbf {b + x} }\right)\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of pushforward measure
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \prod_{i \mathop = 1}^n \left({\left({b_i + x_i}\right) - \left({a_i + x_i}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Lebesgue measure
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \prod_{i \mathop = 1}^n \left({b_i - a_i}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lambda^n \left({\left[\left[{\mathbf a \,.\,.\, \mathbf b}\right)\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Lebesgue measure

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:

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

$\blacksquare$

Note

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

Sources

• René L. Schilling: Measures, Integrals and Martingales (2005)... (previous)... (next) $4.9 \ \text{(i)}$
• René L. Schilling: Measures, Integrals and Martingales (2005)... (previous)... (next) $5.8 \ \text{(i)}$