Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 27G / a

A measure $\mu$ on $(\Omega ,\mathcal{F})$ is $\sigma$-finite whenThe $\sigma$-finite uniqueness theorem says that two measures agreeing on a generating $\pi$-system agree on the generated $\sigma$-algebra, provided the space is covered by countably many sets in that system having finite common measure. Equivalently, a $\sigma$-finite premeasure has at most one extension to the generated $\sigma$-algebra.

Fix $x\in \mathbb{R}$ and define ${\mu }_x(B)=\lambda (B+x)$. This is a $\sigma$-finite measure on the Borel sets. For every half-open interval,The half-open intervals form a generating $\pi$-system, so the $\sigma$-finite uniqueness theorem gives ${\mu }_x=\lambda$. Thus Lebesgue measure is translation invariant.

Let $\mu$ have the stated properties. Translation invariance and the disjoint partition of $(0,1]$ into $n$ intervals of length $1/n$ giveFinite additivity then gives $\mu ((0,r])=r$ for every positive rational $r$. Continuity from below and above extends this to every real $r>0$, and translation invariance yieldsHence $\mu$ and $\lambda$ agree on the half-open intervals. Both are $\sigma$-finite, so uniqueness of extension gives $\mu =\lambda$ on the Borel $\sigma$-algebra.