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$ \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 $\alpha (A) = \beta (A).$

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Some of the hypothesis seems to be missing. Are you sure that you’ve stated the question correctly? –  Brian M. Scott Sep 27 '12 at 5:38
    
The 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. –  tear_drops Sep 27 '12 at 5:40
    
It 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 $A\in\mathscr{F}$? –  Brian M. Scott Sep 27 '12 at 5:43
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No, it isn't correct, consider p. e. $\lambda$ and $0$ on $(\mathbb R, \operatorname{Bor})$. –  martini Sep 27 '12 at 5:43
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@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. –  Brian M. Scott Sep 27 '12 at 6:11

1 Answer 1

Here is a true statement.

Let $ \alpha $ and $\beta $ denote two probability measures on $ (\Omega, \mathscr F) $. If $\alpha (A) \geqslant \beta (A) $ for every $A$ in $\mathscr F$, then $\alpha = \beta$.

Note the added hypothesis that $\alpha$ and $\beta$ are probability measures and the modified hypothesis on the comparison.

To prove this, assume that there exists $A$ in $\mathscr F$ such that $\alpha(A)\ne\beta(A)$. Then $\alpha(A)\gt\beta(A)$, $\alpha(\Omega\setminus A)\geqslant\beta(\Omega\setminus A)$ and $A$ and $\Omega\setminus A$ are disjoint with union $\Omega$ hence $$ 1=\alpha(\Omega)=\alpha(A)+\alpha(\Omega\setminus A)\gt\beta(A)+\beta(\Omega\setminus A)=\beta(\Omega)=1, $$ which is absurd. Thus, $\alpha(A)=\beta(A)$ for every $A$ in $\mathscr F$, that is, $\alpha = \beta$.

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