Thought about this problem a little more carefully, and I believe I have a proper proof.

Suppose that is a Lebesgue measureable set. Then which is to say and since is arbitrary, we must have that . Note that is an open set such that (but becomes equal to E in the limit represented by the infimum!).

Correspondingly, we have by the above definitions: . But the term is finite, and so can . It follows is measurable.

To prove its measure is as we expect, we make the following computation:

.

The justifications at various steps follow from the related definitions (dilation sets), hypothesis of Lebesgue measure (i.e. being able to pass to open epsilon-close covers, and thus to the limit represented as an inf), and the open set decomposition theorem (i.e. being able to then pass from open coverings to countable coverings of cubes).

Please let me know if there are any flaws in my argument, and how they can be fixed! I'm still a little uneasy about the level of rigor in my solution, particularly in passing from the set in question, to open coverings, to countable unions of cubes and the use of the infimum along the way.