2 simple outer measure problems

Hello all.

These are homework problems, so partial solutions / advice would be preferred.

These come from the book Real Analysis (4th ed) by Royden and Fitzpatrick

Chapter 2 problems 15 and 20

15. Show that if E has finite measure and $\displaystyle \epsilon>0$, then E is the disjoint union of a finite number of measurable sets, each of which has measure at most $\displaystyle \epsilon$.

[For this one I don't really know how to begin. At this point we know

Theorem 9: The collection M of measurable sets is a $\displaystyle \sigma$-algebra that contains the $\displaystyle \sigma$-algebra B of Borel sets. Each interval, each open set, each closed set, each $\displaystyle G_\delta$ and each $\displaystyle F_\sigma$ set is measurable.

If I'm not mistaken this means that once we prove problem 15, we will have also shown that any set with finite outer measure is measurable (since it is a finite union of measurable sets)

Thoughts? ]

20.(Lebesgue) Let E have finite outer measure. Show that if E is measurable if and only if for each open, bounded interval (a,b),

b-a=m*($\displaystyle (a,b) \cap E$)+m*((a,b)~E)

[Here one direction is very easy. If E is measurable, just let (a,b) be a testing set and we're done. As for the other direction...?

In this section we learned the excision property m*(B~A)=m*(B)-m*(A) and

Theorem 11: The following are equivalent

0)E measurable

1) $\displaystyle \exists$ Outer Approximation by Open Sets and $\displaystyle G_\delta$ Sets

2) $\displaystyle \exists$ Inner Approximation by Closed Sets and $\displaystyle F_\sigma$ Sets

and

Theorem 12: Let E be a measurable set of finite outer measure. Then for each $\displaystyle \epsilong>0$, there is a disjoint collection of open intervals $\displaystyle {{\{I_k\}}_{k=1}}^n$ for which if O=union of $\displaystyle I_k$, then m*(E~O)+m*(O~E)$\displaystyle <\epsilon$

I don't really know how to go from making a statement about all open intervals to a statement about all sets of real numbers.

Thoughts?]

Thank you all.