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?


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.

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

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