$ \alpha $ and $\beta $ are measures on $ (\Omega, \mathscr F) $ and $ A \subset \mathscr F$. If $\alpha (A) \ge \beta (A) $, I need to prove that $\alpha (A) = \beta (A).$
If $\alpha (A) \ge \beta (A) $ then $\alpha (A) = \beta (A)$, where $ \alpha $ and $\beta $ are measures
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measure-theory
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0Some of the hypothesis seems to be missing. Are you sure that you’ve stated the question correctly? – 2012-09-27
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0The question is correct according to my notes. I am wondering about how to prove it, so just added it here to get the experts' feedback. – 2012-09-27
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0It simply can’t be proved on the basis of the information given. Are these probability measures, and does the hypothesis $\alpha(A)\ge\beta(A)$ hold for **all** $A\in\mathscr{F}$? – 2012-09-27
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1No, it isn't correct, consider p. e. $\lambda$ and $0$ on $(\mathbb R, \operatorname{Bor})$. – 2012-09-27
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0@martini, Please explain more. – 2012-09-27
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2@martini simply gave an example of two measures, $\lambda$ and $0$, on the field of Borel subsets of $\Bbb R$ that satisfy the condition $\lambda(A)\ge 0(A)$ for all $A$ in the field, yet $\lambda(A)\ne 0(A)$ for a great many $A$ in the field. This shows that the result is simply false without further hypotheses. – 2012-09-27
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0By $A\subset\mathscr F$ do you mean $A\in\mathscr F$? – 2016-09-15