# Smallest set of a full measure

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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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

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