# theorem statement

• Nov 12th 2013, 08:38 PM
director
theorem statement
Given the following theorem:

Quote:

Let E by any set of real numbers. The following assertion is equivalent to the measurability of E:
For each epsilon>0, there is an open set O containing E for which m*(O~E)<epsilon
(where m*() denotes Lebesgue outer measure)
Is there any way I can manipulate this statement to get the condition that occurs when E is not measurable?

Thanks
• Nov 12th 2013, 08:57 PM
SlipEternal
Re: theorem statement
Negate the statement.

Let $\displaystyle E\subseteq \mathbb{R}$ be any non-measurable set. Then there exists $\displaystyle \varepsilon>0$ such that for all open sets $\displaystyle O$ with $\displaystyle E\subseteq O$, $\displaystyle m^*(O\setminus E)\ge \varepsilon$.

A theorem that is equivalent to a definition implies a biconditional relationship. The set is measurable if and only if the theorem applies, so if you negate the statement of the theorem, you get the opposite result.
• Nov 13th 2013, 04:22 AM
director
Re: theorem statement
That's sort of what I was thinking, thank you!