Let $P$ and $Q$ be two probability measures on a measure space $(\Omega,\mathcal{F})$. Show that if $P(A) = Q(A)$ for all $A\in \mathcal{F}$ such that $P(A) \leq 1/2$, then $P=Q$. Show that the result fails if we replace the constraint $P(A) \leq 1/2$ by the more stringent $P(A) < 1/2$.

Is my proof for part 1 correct? Also I would like some tips/ideas for how to approach part 2


Suppose $P$ and $Q$ are two probability measures on a measurable space such that $P(A) = Q(A)$ whenever $P(A) \leq 1/2$. We want to show $P$ and $Q$ agree on all $A\in \mathcal{F}$ such that $P(A)>1/2$. Let $A\in \mathcal{F}$ such that $P(A)> 1/2$. then $$ P(A^C) = 1 - P(A) < 1/2 $$ Therefore $P(A^C) = Q(A^C)$. Since we also have $Q(A^C) = 1-Q(A)$, then $1-Q(A) = 1-P(A)$ implies $Q(A)=P(A)$.



1 Answer 1


Your proof for the first part seems fine.

Hint for the second part: Note that $P$ and $Q$ can only possibly disagree on a set $A$ with $P(A) = \frac{1}{2}$. Furthermore, it would be nice if $P$ does not take any values smaller than $\frac12$ other than $0$.

Does this help?

  • $\begingroup$ I ended up just saying: Consider the sigma-algebra: $\{\emptyset, A, A^C, \Omega\}$, with $P(\emptyset)=0=Q(\emptyset)$, $P(\Omega)=1=Q(\Omega)$, but $P(A) = 1/2$ and $Q(A)=1/4$. $\endgroup$
    – bdeonovic
    Oct 6, 2015 at 12:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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