# Finite OUTER measure and measurable set

Royden's Real Analysis (4th edition), problem #19 (Chapter 2.5):

Let $E$ have a finite OUTER measure. Show that if $E$ is not measurable, then there is an open set $O$ containing $E$ that has a finite outer measure and for which $m^{*}(O - E) > m^*(O)-m^*(E)$.

My question is how can a set of finite measure be not measurable? I know that every set of finite positive measure harbors non-measurable subsets, but how could the whole set $E$ be not measurable when it has a finite measure by assumption?

Thanks.

I have righted the above problem. Sorry all!

• It makes no sense. Something is missing for sure. Oct 16, 2015 at 3:24
• $E$ has finite measure and then we should consider if $E$ is not measurable??? I think you meant: "Let $E$ have finite outer measure. ...". Oct 16, 2015 at 17:47

It's probably supposed to say "Let $E$ have finite outer measure".

There's an errata list here. There is no entry for this problem, but the entry for problem 18 on page 43 looks similar to this problem and is supposed to start "Let $E$ have finite outer measure".

• My mistake. I misread: it does say finite "outer" measure. Sorry! Oct 16, 2015 at 23:23

Since $E$ is not measurable, we know by Theorem 11 (Royden 4th edition) that there exists an $\epsilon > 0$ such that for any open set $O$ containing $E$ we have $m^∗ (O - E) ≥ \epsilon$. By the definition of outer measure, we know that there exists a countable collection of bounded open intervals $\{I_k\}_{k=1}^{\infty}$ whose union we denote $O ≡ \bigcup_{k=1}^{\infty} I_k$, such that $m^{∗} (O) − m^∗ (E) < \epsilon \leq m^∗ (O - E)$.

The source of the solution given above is found here.

• Idon't understand the last line. By the definition of outer measure, we know that there exists a countable collection of bounded open intervals $\{I_k\}_{k=1}^{\infty}$ whose union we denote $O ≡ \bigcup_{k=1}^{\infty} I_k$, such that $m^{∗} (O) − m^∗ (E) < \epsilon$. I don't see that in the def'n of outer measure in the book. Oct 17, 2016 at 16:45
• @NinosławCiszewski Since the outer measure is an infimum (i.e. greatest lower bound) then there must exist an open interval cover with sum of lengths $\sum_{k=1}^{\infty}{\ell(I_k)} < m^*(E) + \epsilon$ for all $\epsilon > 0$ since otherwise $m^*(E) + \epsilon$ would be a greater lower bound than $m^*(E)$ which would be a contradiction. Combined with subadditivity of outer measure $m^*(E) + \epsilon > \sum_{k=1}^{\infty}{\ell(I_k)} \geq m^*\left( \bigcup_{k=1}^{\infty} I_k \right)$ Just let $O = \bigcup_{k=1}^{\infty} I_k$ and you have the result about. Sep 9, 2018 at 22:02

Consider using a dyadic mesh argument to construct a family of open sets each of which contains $E$, and then take their intersection.