Talk:Singular measure

mutually singular
so why not call this article "mutually singular measures" like the first line says? --itaj 04:01, 12 May 2006 (UTC)

a question
let $$(\Omega,\Sigma),$$ be a measurable space. i'm talking here about real-value signed finite measures on this space. let M be a set of measures.

let N be the set of all measures absolutely continuous with respect to a measure in the linear span of measures in M. i.e. $$N := \{ \nu : \exists \mu\in span(M)\ (\nu<<\mu) \} $$

for a measure $$\mu$$ i'll denote $$ {singl}(\mu) := \{ \nu : \nu\perp\mu \} $$. the set of measures mutually singular to $$\mu$$.

and for a set of measures L. $$ {Singl}(L) := \{ \nu : \forall \mu\in L\ (\mu\perp\nu) \} $$. the set of all measures mutually singular to all measures in L.

my question is if the above implies that $$ N = Singl(Singl(N)) $$ i know this is true if there's only one measure in M, but i need to know about infinite set M, countable and bigger. --itaj 04:05, 12 May 2006 (UTC)


 * If span is in the sense of vector spaces, then no. Consider $$M:=\{\delta_n:n\in N\}$$.  Then $$\sum 2^{-n}\delta_n\not\in span(M)$$ or N, but it is in singl(singl(N)).  (Cj67 23:39, 25 June 2006 (UTC))