Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to ask if it is true in general topological space that the countable union of sets of measure $0$ has $0$ measure?

share|cite|improve this question
It is true in general measure spaces as a consequence of countable additivity. – Michael Greinecker Apr 15 '12 at 21:43
On the other hand, a countable union of meager sets is meager, in any topological space (meager means it is a countable union of nowhere dense subsets). Meager sets are the usual notion of "negligible" in the context of topological spaces, while "contained in a set of measure zero" is the usual notion of "negligible" in measure spaces. – Arturo Magidin Apr 15 '12 at 23:19

A general topological space doesn't have a notion of "measure 0".

The statement you ask about is true in a general measure space. By definition, measures are required to be countably additive. This means that $$ m\left(\bigcup_{i=1}^\infty S_i\right) \;=\; \sum_{i=1}^\infty m(S_i) $$ for any countable disjoint collection of measurable sets $\{S_i\}_{i=1}^\infty$. As a consequence, if $\{S_i\}_{i=1}^\infty$ is any countable collection of measurable sets (not necessarily disjoint), then $$ m\left(\bigcup_{i=1}^\infty S_i\right) \;\leq\; \sum_{i=1}^\infty m(S_i) $$ In particular, if each $S_i$ has measure $0$, then the union must also have measure $0$.

share|cite|improve this answer

"Measure" doesn't make sense in a general topological space. In a general measure space, a countable union of measure-zero sets always has measure zero, by the countable subadditivity of measures.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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