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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

PART 1

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 $Q(A)=P(A)$.

PART 2

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1 Answer 1

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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?

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  • $\begingroup$ 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

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