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Let $\lambda $ be a positive or complex measure. We say that $\lambda $ is concentrated on $A $ if for some set $A \in \mathcal{B } $ we have that $\lambda (E) =\lambda (A \cap E ) $ for every $E \in \mathcal {B }$ , where $\mathcal {B } $ is some $\sigma $-algebra.

I want to show that this is equivalent to the hypothesis that $\lambda (E) =0 $ whenever $E \cap A = \emptyset $.

Thanks in advance!

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  • $\begingroup$ Tried something? $\endgroup$ – Did Oct 3 '14 at 9:05
  • $\begingroup$ Yes, I can sort of "see" the equivalence of these two definitions. But I don't know how to formulate a proof. $\endgroup$ – Alexander Oct 3 '14 at 9:19
  • $\begingroup$ Ok, I did it myself. $\endgroup$ – Alexander Oct 3 '14 at 10:04
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Suppose $\lambda (E) =0 $ whenever $A \cap E = \emptyset $. Now let $F $ be any set, then $F = (F \cap A) \cup ( F \cap A ^c )$, and since this is a disjoint union and $(F \cap A^c)\cap A = \emptyset $ so that $\lambda(F \cap A^c)=0 $. We then have that $\lambda (F)=\lambda (F \cap A ) $.

Conversely, suppose $\lambda (E)=\lambda(E \cap A)$, then if $E \cap A=\emptyset $, $\lambda(E)=\lambda(\emptyset)=0 $.

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