Question 18 of Section 2.5 (page 43) in this link : http://math.harvard.edu/~ctm/home/text/books/royden-fitzpatrick/royden-fitzpatrick.pdf claims that if a set $ E $ has finite outer measure, then, there is a set $ F \in F_{ \delta} $ such that, we have $ F \subset E $ and $ m^{*}(F) = m ^ { * } (E) $.
However, I am able to prove that I can find such an $ F $ only if $ E $ is measurable. Can someone figure out if there is any mistake in the proof below? Or is the exercise an oversight?
Proof. Suppose that there is an $ F \in F_{ \delta } $ such that $ m ^{ * } ( F) = m ^{ * } ( E ) $. Then, we apply the excision property (page 40 of Royden) on the measurable set $ F $, which is of finite measure, to deduce that $ m ^ { * } ( E - F ) = m ^ { * } ( E ) - m ^{*} (F) = 0 $. Then, by Theorem 11 (iv) (page 40), we deduce that $ E $ is measurable.
This is in the errata (cf. http://www2.math.umd.edu/~pmf/err130405.pdf); indeed the desired condition is actually equivalent to measurability. The statement can be rephrased (in terminology not used in Royden and Fitzpatrick) as saying that the outer and inner measures of $E$ are the same.