I have to prove that if a set $E$ that is Jordan-measurable, then $E$ is measurable in the sense of Lebesgue.
In order to do this, I am required to use that a set is Jordan-measurable iff $bd(E)$ has content (or measure, in this case) zero. (Here $bd$ denotes de boundary of $E$).
I have to find an open set $G$ and closed $F$ such that $F\subset A \subset G$ and $\mu(G-F)<\epsilon$, but I don't know how to start… Any idea?
Best Answer
(Here comes my first post!)
Since the interior of E is an open set and the closure of E is a closed set, then both of them are Lebesgue Measurable (why?).
Then prove the fact that the measure of the closure of E is equal to the measure of the interior of E (you dont need this explicitly but its enlightening). From there, and the fact that the interior is a subset of E and E is a subset of its closure, plus the definition of Lebesgue Measure for both the interior and the closure, should ring lots of bells to you.
Keep in mind the relationship between the boundary of E, its interior and its closure. Also monotony and sub-additivity of outer measure come in handy, hope it helps!