Let X be an arbitrary space, and let μ, ν be two probability measures on X. Suppose there is a common set A on which μ ( A ) = ν ( A ) = 1.Now suppose there are B , C such that μ ( B ) = ν ( C ) = 1. Can anything be said of μ ( C ) and ν ( B ), or μ ( B ∩ C ) and ν ( B ∩ C )?My initial intuition was that the answer should be yes since B ∩ A and C ∩ A both have full μ and full ν measure respectively, and so there should be "enough" overlap for B ∩ C to have at least positive measure. But I can't seem to prove this - so I'm interested in whether anything can be said of these quantities (either alone or possibly together, e.g. is it possible for μ ( C ) = ν ( B ) = 0) and whether there is any clever counterexample for which μ ( C ) and ν ( B ) can be any arbitrary number in [0,1].

Question

Let   X be an arbitrary space, and let   μ,  ν be two probability measures on   X. Suppose there is a common set   A on which   μ  (  A  )  =  ν  (  A  )  =  1.Now suppose there are   B  ,  C such that   μ  (  B  )  =  ν  (  C  )  =  1. Can anything be said of   μ  (  C  ) and   ν  (  B  ), or   μ  (  B  ∩  C  ) and   ν  (  B  ∩  C  )?My initial intuition was that the answer should be yes since   B  ∩  A and   C  ∩  A both have full   μ and full   ν measure respectively, and so there should be &quot;enough&quot; overlap for   B  ∩  C to have at least positive measure. But I can&#039;t seem to prove this - so I&#039;m interested in whether anything can be said of these quantities (either alone or possibly together, e.g. is it possible for   μ  (  C  )  =  ν  (  B  )  =  0) and whether there is any clever counterexample for which   μ  (  C  ) and   ν  (  B  ) can be any arbitrary number in [0,1].

Braedon Rivas · Accepted Answer

The problem is that such a set   A always exists, namely   X itself. Hence, we can&#039;t really say anything non-trivial.

fabios3 · Accepted Answer

If   A  ⊊  X, I won&#039;t be able to say anything, since   A can &quot;almost&quot; be   X

Let X be an arbitrary space, and let μ , ν be two

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