# Why does an open interval NOT have measure zero?

I am currently working on a proof that requires me to show that an open ball $B_{\epsilon}(x)$ has nonzero measure. I currently have the following proof in my book:

"The closed interval $[a,b]$ is not of measure zero."

Hence, if I take the contrapositive, does it follow that "if an interval doesn't have measure zero, the interval is open"? Is there a way for me to prove that the open interval on the line $\mathbb{R}$ is of measure not zero? Thanks!

• The contrapositive you give is not correct. It should be "If a set is of measure zero, then it is not a closed interval." (Of course, it is assumed that a closed interval $[a,b]$ is such that $a\neq b$, as otherwise this would not be true.) – Hayden May 18 '14 at 1:59
• Closed and open are not polar opposites. You can have clopen sets in addition to open and closed sets. – IAmNoOne May 18 '14 at 2:06
• And, in addition to open, closed, and clopen sets, you can have ones which are neither (which is perhaps more relevant here, since there are no clopen sets in $\mathbb{R}$, but a set like $(0,1]$ is neither closed nor open) – Milo Brandt Oct 30 '14 at 2:10

The closed interval $[a,b]$ has measure $0$ when $a\ge b$, so this was presumably in a context where one knows that $a<b$.
A sort of contrapositive would say that if an interval does have measure $0$, then it is not a closed interval $[a,b]$ where $a<b$. That certainly does not mean it is an open interval.
It is clearly false that if an interval does not have measure $0$, then it is open. Any interval $[a,b]$ where $a<b$ is a counterexample to that statement. Moreover, if you do show that every open interval has positive measure, that's not at all the same as showing that if some assumption holds, then an interval is open.
Your contrapositive seems to be wrong in two ways. An interval not being closed $[a,b]$ does not mean it is open, and the negation of "not of measure zero" should be "of measure zero". In any case the statement is missing the assumption $a<b$, at least for the Lebesgue measure.