Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
No, it isn't correct, consider p. e. $\lambda$ and $0$ on $(\mathbb R, \operatorname{Bor})$. – martini Sep 27 '12 at 5:43
@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

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, $\alpha = \beta$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.