1
$\begingroup$

A measure space $(X,F,\mu)$ is complete if whenever $\mu(E)=0$, every subset of $E$ is an element of $F$. Every incomplete measure space has a completion, which extends the sigma algebra to more sets and extends the measure to those sets. Most famously, the Borel algebra on $\mathbb{R}$ is incomplete with respect to Lebesgue measure, but it can be extended to a bigger sigma algebra, the Lebesgue signma algebra, which is complete with respect to Lebesgue measure.

But question is, does there exist a measure under which the Borel sigma algebra on $\mathbb{R}$ is complete? Or is it incomplete under every measure?

$\endgroup$
3
$\begingroup$

The Borel $\sigma$-algebra on any space is complete with respect to counting measure, since the only null set for counting measure is the empty set.

However, the Borel $\sigma$-algebra on $\mathbb{R}$ is not complete with respect to any $\sigma$-finite measure. Note that $\mathbb{R}$ is Borel-isomorphic to $\mathbb{R}^2$ (see Are the measurable spaces $(\mathbb{R}^n, Bor(\mathbb{R}^n))$ and $(\mathbb{R}^m, Bor(\mathbb{R}^m))$ isomorphic for $n\neq m$, for instance), so let us work with $\mathbb{R}^2$ instead of $\mathbb{R}$. So suppose $\mu$ is a $\sigma$-finite Borel measure on $\mathbb{R}^2$. Note then the sets $\mathbb{R}\times\{t\}$ as $t$ ranges over $\mathbb{R}$ are disjoint, and so by $\sigma$-finiteness of $\mu$ we must have $\mu(\mathbb{R}\times\{t\})=0$ for all but countably many $t\in\mathbb{R}$. Since not every subset of $\mathbb{R}\times\{t\}$ is Borel, this measure space is not complete.

$\endgroup$
  • $\begingroup$ Nice answer @Wofsey . $\endgroup$ – Kavi Rama Murthy Oct 25 '18 at 5:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.