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.

  • 1
    $\begingroup$ 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 $\mathcal{A}$. $\endgroup$ Jan 22, 2020 at 13:11
  • $\begingroup$ @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? $\endgroup$
    – ChanaG
    Jan 22, 2020 at 13:22
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    $\begingroup$ 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. $\endgroup$ Jan 22, 2020 at 13:25


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