# A Borel measure s.t. $\mu([a,b])=b-a$ is translation invariant

Let $$\mu$$ be a Borel measure on $$\mathbb{R}$$, i.e. a measure defined on the Borel $$\sigma$$-algebra of $$\mathbb{R}$$. Suppose $$\mu([a,b])=b-a$$ for any closed interval.

Consider the collection of sets $$\mathcal{A}=\{E|\forall x:\mu(E)=\mu(x+E)\}$$, i.e. the colleciton of sets for which the measure $$\mu$$ is translation invariant.

Show that $$\mathcal{A}$$ is a $$\sigma$$-algebra. (As $$\mathcal{A}$$ includes all closed intervals, this is actually the Borel $$\sigma$$-algebra).

It is easy to show that $$\emptyset$$,$$\mathbb{R}$$ are in $$\mathcal{A}$$ , that $$\mathcal{A}$$ is closed under taking complements and closed under countable disjoint union. The missing link is showing that $$\mathcal{A}$$ is closed under finite union (not necessarily disjoint). Any ideas?

P.S: I know that a Borel measure that gives any interval its length is identical to the (restriction of) Lebesgue measure and hence translation invariant, but I want an elementary proof not using this fact.

• So you've shown that $\mathcal{A}$ is a $\lambda$-system contained in the Borel $\sigma$-algebra. Thus it would suffice to find a suitable $\pi$-system contained in $\mathcal{A}$. Jan 22, 2020 at 13:11
• @DanielFischer - Elegant solution, thanks! Just to make sure I got you right - the collection of closed intervals (and the empty set) is a suitable $\pi$-system? Jan 22, 2020 at 13:22
• Yes. It is a $\pi$-system generating the Borel $\sigma$-algebra, hence $\mathcal{B}\subset \mathcal{A}$ by the $\pi$–$\lambda$-theorem, and $\mathcal{A}\subset \mathcal{B}$ trivially. Jan 22, 2020 at 13:25