A Wikipedia page on filters claims:

if $(X, \mu)$ s a measure space, then the collection $\{A : \mu(A) > 0\}$ is a filter.

I assume that the implied poset is the $\sigma$-algebra on $X$ partially ordered by set inclusion, so $A$ in the above definition is a measurable subset of $X$.

Assuming that at least one set with positive measure exists, this collection clearly satisfies two filter axioms, namely nontriviality and upward closure. However, I'm confused with downward directedness. Following the axiom:

For every $x, y \in F$, there is some $z \in F$ such that $z \leq x$ and $z \leq y$.

If we take some non-intersecting $A$ and $B$ with positive measure, they will be in the filter. But there is no $C$ of positive measure that is included in both $A$ and $B$.

Am I missing something, or is this example incorrect?

  • $\begingroup$ I think for the positive measure example, $\leq$ means $\supseteq$, so for $x=A$ and $y=B$ one can use $z = A \cup B.$ Unless I'm overlooking something, the Wikipedia article at this location needs to be more explicit. FYI, I never liked using $\leq$ this way for filters, as I think of filters as collections of BIG things, and bigger (i.e. $\geq)$ than big is big. $\endgroup$ Sep 24 at 19:16
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    $\begingroup$ @DaveL.Renfro this does not work, Think about a measure on subsets of X such that X has a positive measure. Because X is the ≤-minimum, you would get that all subsets of X are in the that said filter, hence all (measurable) subsets of X are of positive measure $\endgroup$
    – ℋolo
    Sep 24 at 19:31
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    $\begingroup$ The example is incorrect. I think that what happened is that someone remembered that "the set of null-sets (measure 0 sets)" form an ideal, and because ideal is the dual of a filter they wanted to say (correctly) that "the dual of the ideal of null-sets is a filter", but they translated this "dual" incorrectly, the correct dual of that ideal is {A | μ(X\A)=0}, and the correct statement is "{A | μ(X\A)=0} is a filter" $\endgroup$
    – ℋolo
    Sep 24 at 19:34
  • $\begingroup$ @ℋolo: I often get switched up with filters because I'm much more used to dealing with ideals, and for ideals it's the union that is small, so for filters it is the intersection that is big. Probably what was intended in the Wikipedia article at that location is the complement of finite-measure sets, although complements of zero-measure sets (what you gave) also serves as an example (in this case, it is the dual of a $\sigma$-ideal, not just the dual of an ideal, which is why I suspect they meant complements of finite-measure sets). $\endgroup$ Sep 24 at 19:47
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    $\begingroup$ @DaveL.Renfro they definitely didn't mean the complete finite measure sets because they gave it as the next example in the wiki page $\endgroup$
    – ℋolo
    Sep 24 at 19:51

1 Answer 1


You are correct, the example is (was) incorrect.

My guess is that someone tried to explicitly describe the dual of the null-sets ideal but did it incorrectly.

I have edited the wiki page to have the correct definition of the dual filter of the null-sets ideal (here is the old revision)


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