Here's my unoriginal attempt:Show that if a set X in R has positive outer measure, then there is a bounded subset of X also having positive outer measure.

Let X be any set in R, X measurable, $\displaystyle 0 < m*(X) < \infty$, so X is bounded

Let $\displaystyle Y = X \cap Q$ and $\displaystyle Y' = X \sim Y$

So, Y, Y' bounded

$\displaystyle m*(Y) = 0$ (countable set)

$\displaystyle m*(X) = m*(Y \cup Y') = m*(Y) + m*(Y') = 0 + m*(Y')$

So, $\displaystyle m*(Y') = m*(X) > 0$

Does this make sense?

Another way I wanted to show this was by representing X as a disjoint union of (finite number of) measurable, bounded sets. Then one of these sets would satisfy the criteria, I think.

If X open, I know that X is the disjoint union of a countable collection of open intervals.

If X closed, or X neither open or closed - what is the disjoint union?

Thanks for any tips.