In a Lemma to be used in developing properties of Lebesgue Outer Measure on $R^N$ ($N$ is a positive integer) the professor pointed that it is not hard to establish that the volume of an interval $I$ in $R^N$ is equal to the supremum of the following set {volume of $K$|$K$ is a compact interval in $R^N$ contained in $I$}.
Now one way of the inequality is clear which is that the volume of $I$ is greater than the least upper bound of that set. For the other inequality if $\epsilon>0$ is given I tried to prove the existence of an interval $K_\epsilon$ whose volume is greater than the volume of $I$ $-\epsilon$ but I don't seem to know how to give this precisely. Any hint or help is appreciated.
The Lemma I'm talking about states that the volume of an interval is less than the sum of volumes of any countable covering by intervals.
By interval, $I$, assume you mean $\prod (a_i, b_i)$, where $a_i,b_i \in [-\infty, +\infty]$. If either $a_i$ or $b_i \in \{-\infty,+\infty\}$, then $vol(I) =\infty$, and its not hard to find compact interval $K\subset I$ with arb large volume. If all the $a_i,b_i$ are finite and $\delta$ is small enough, if $K=\prod [a_i+\delta,b_i-\delta]$, $vol(K)$ can be made to be within $\epsilon$ of $vol(I)$.