Lebesgue Measure is Invariant under Translations

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Theorem

Let $\lambda^n$ be the $n$-dimensional Lebesgue measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\map \BB {\R^n}$.

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


Then $\lambda^n$ is translation-invariant; that is, for all $B \in \map \BB {\R^n}$, have:

$\map {\lambda^n} {\mathbf x + B} = \map {\lambda^n} B$

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


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 $\map \BB {\R^n} \, / \, \map \BB {\R^n}$-measurable.

Consider the pushforward measure $\lambda^n_{\mathbf x} := \paren {\tau_{\mathbf x} }_* \lambda^n$ on $\map \BB {\R^n}$.


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

$\map \BB {\R^n} = \map \sigma {\JJ^n_{ho} }$

where $\sigma$ denotes generated $\sigma$-algebra, and $\JJ^n_{ho}$ 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:

$J_k := \hointr {-k} k^n$


For condition $(3)$, let $\horectr {\mathbf a} {\mathbf b} \in \JJ^n_{ho}$ be a half-open $n$-rectangle.

Since:

$\map {\tau_{\mathbf x}^{-1} } {\horectr {\mathbf a} {\mathbf b} } = \mathbf x + \horectr {\mathbf a} {\mathbf b} = \horectr {\mathbf {a + x} } {\mathbf {b + x} }$

we have:

\(\ds \map {\lambda^n_{\mathbf x} } {\horectr {\mathbf a} {\mathbf b} }\) \(=\) \(\ds \map {\lambda^n} {\horectr {\mathbf {a + x} } {\mathbf {b + x} } }\) Definition of Pushforward Measure
\(\ds \) \(=\) \(\ds \prod_{i \mathop = 1}^n \paren {\paren {b_i + x_i} - \paren {a_i + x_i} }\) Definition of Lebesgue Measure
\(\ds \) \(=\) \(\ds \prod_{i \mathop = 1}^n \paren {b_i - a_i}\)
\(\ds \) \(=\) \(\ds \map {\lambda^n} {\horectr {\mathbf a} {\mathbf b} }\) Definition of Lebesgue Measure

Finally, since:

$\ds \map {\lambda^n} {J_k} = \prod_{i \mathop = 1}^n \paren {k - \paren {-k} } = \paren {2 k}^n$

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


Whence Uniqueness of Measures implies that:

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

and since for all $B \in \map \BB {\R^n}$ we have:

$\mathbf x + B = \map {\tau_{\mathbf x}^{-1} } B$

this precisely boils down to:

$\map {\lambda^n} {\mathbf x + B} = \map {\lambda^n} B$

$\blacksquare$


Motivation

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


Sources