Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Let $(X,\mathscr F,\mu)$ be a measurable space with a positive measure $\mu$. Let us call the set $F\in\mathscr F$ full if $\mu(F^c) = 0$. I wonder if there exists the smallest full set $F\in \mathscr F$ in the following sense:

  1. if $B\supseteq F$, $B\in \mathscr F$ then $\mu(B^c) = 0$;

  2. if $F'\in \mathscr F$ is another set satisfying 1. then $\mu(F\setminus F') = 0$, i.e. $F$ lies in $F'$ almost completely.

If the existence does not hold for the general setting, would it be sufficient to assume that $\mu$ is $\sigma$-finite and/or that $\mathscr F$ is countably generated?

share|cite|improve this question
It seems to me that any full set satisfies the two conditions that you give, so this isn't a very useful definition of "smallest full set". – Jim Belk May 3 '12 at 13:11
@Jim: you're certainly right, thank you – Ilya May 3 '12 at 13:17
A related (distantly) question. Suppose you have a Borel measure on a topological space. If you take the union of all open sets of measure zero, must that be a set of measure zero? The complement of that union would be called the "support" of the measure. For some spaces, yes (for example separable metric spaces), but for others, no. – GEdgar May 3 '12 at 14:32
@GEdgar: Thanks for the comment. I was aware of that support of the measure - but I didn't want to use any additional structure on $X$ to define the smallest full set. Michael has shown and Jim has mentioned that the definition I gave includes all full sets, hence not useful. W.r.t. support, in fact, Byron has left the link to that question but then deleted it - so I read that thread and also the linked MO thread. – Ilya May 3 '12 at 14:50
up vote 4 down vote accepted

$X$ itself is always full. If you have a diffuse measure, $F$ is full and $x\in F$, then $F\backslash\{x\}$ is full too, so there will be in general no smallest full set under set inclusion, even for nonpathological measures such as the uniform distribution on $[0,1]$ with the Borel $\sigma$-algebra. You can show that $F$ is full if and only if $\mu(F\Delta X)=0$.

share|cite|improve this answer
Thank you very much - I knew that there is no the smallest full set w.r.t. a set inclusion even in very nice cases - that's why I've asked in 2. for the almost inclusion. But that appeared to be wide enough to include all full sets as Jim has stricken out in his comment. – Ilya May 3 '12 at 14:56

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.