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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?

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    @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](http://math.stackexchange.com/questions/115078/is-the-support-of-a-borel-measure-measured-the-same-as-the-whole-space) but then deleted it - so I read that thread and also the linked MO thread.2012-05-03

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$X$ 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 $\mu(F\Delta X)=0$.

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    Th$a$nk 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.2012-05-03