6

I was wondering if somebody could help me out with a solution for the following problem (taken from Royden's Real Analysis, 4e. (ch. 2.4, prob. 18):

Let $E$ have finite outer measure. Show that there is an $F_\sigma$ set $F$ and a $G_\delta$ set $G$ such that $F \subset E \subset G$ and $m^*[F]=m^*[E]=m^*[G]$.

So far I constructed a candidate set $G$ (for each $n \in \mathbb{N}$ there is a countable open interval cover $\{ I_{n,k} \}$ such that $\Sigma_k \ \ell(I_{n,k}) < m^*[E] + \frac{1}{n}$; then set $G:=\cap_n \cup_k I_{n,k}$). Since $G$ is measurable, I can approximate it with an $F_\sigma$ set $F \subset G$ such that $m^*[F]=m^*[G]$, but of course there is no guarantee that $F \subset E$. Is there anyway to salvage this proof by toying with $F$ so that $F \subset E$ but still remains an $F_\sigma$ set? Or should I approach another way?

Thanks in advance for any help!

user121410
  • 359
  • 4
  • 10
  • 1
    Just to add to the earlier answer provided by @DanielFIscher (which is correct), this question as stated in the textbook is listed as an error in the errata list published online (just search for errata Royden Fitzpatrick). The corrected version now appears in the latest version of the book. – user650503 Mar 04 '19 at 06:56

1 Answers1

7

We need a further condition on $E$, $m^{\ast}[E] < \infty$ alone is not sufficient:

Since $F_\sigma$ and $G_\delta$ sets are Lebesgue-measurable (they are Borel sets), if we have an $F_\sigma$ set $F$ and a $G_\delta$ set $G$ with $F \subset G$ and $m^\ast[F] = m^\ast[G] < \infty$, then $G\setminus F$ is a $G_\delta$ set with $m^\ast[G\setminus F] = 0$. Since the Lebesgue measure is complete, every set between $F$ and $G$ is thus Lebesgue-measurable (a union of an $F_\sigma$ with a null set).

Thus measurability of $E$ is a necessary condition for the conclusion to hold.

By the outer regularity, it is also a sufficient condition: Let $K_n = \{ x : \lvert x\rvert \leqslant n\}$. For every $k$, there is an open set $O_k \supset (K_n\setminus E)$ with $m^\ast[O_k] < m^\ast[K_n\setminus E] + 2^{-k}$. Then

$$F_n = \bigcup_k (K_n\setminus O_k)$$

is an $F_\sigma$ set with $F_n \subset E\cap K_n$ and $m^\ast[F_n] = m^\ast[E\cap K_n]$. Further,

$$F = \bigcup_n F_n$$

is an $F_\sigma$ set contained in $E$ with $m^\ast[F] = m^\ast[E]$.

Daniel Fischer
  • 212,535
  • Thanks, Daniel, really helpful! I think I understand now. It turns out I had the statement of the excision property wrong; I assumed you could only excise a measurable set from a measurable set, when in fact you could excise a measurable set from any set (and this helps me finish my proof!). Anyway, thanks again! – user121410 Feb 15 '14 at 16:55
  • Hi @DanielFischer Im trying this problem now and I came across this post. How do we know that $E$ is the union from the above $F_{\sigma}$ set $F$ and a null set $N \subset G\setminus F$. Doesnt this mean you already assumed that $F \subset E \subset G$ with $m^\ast[G\setminus F] = 0$ which we dont know. We only know that $E \subset G$ and that there exists an $F_{\sigma}$ such that $m^\ast[G\setminus F] = 0$ but we dont know that $F \subset E$ – alpastor May 22 '16 at 21:02
  • @NickE I'm not sure what you're referring to. The first paragraph? I should probably have used different letters, and it's not important that we have a $G_{\delta}$ and an $F_{\sigma}$. Point is, if $A,B$ are Lebesgue-measurable, $A \subset B$ and $m^{\ast}[A] = m^{\ast}[B] < \infty$, then every set between $A$ and $B$ is also Lebesgue measurable. Hence the situation $F \subset E \subset G$ with $F$ and $G$ Lebesgue-measurable and $m^{\ast}[F] = m^{\ast}[E] = m^{\ast}[G] < \infty$ can only occur if $E$ is Lebesgue measurable. It is not sufficient that $m^{\ast}[E] < \infty$. – Daniel Fischer May 22 '16 at 21:17
  • @DanielFischer Ohh sorry I misread, you're right. I thought the first paragraph was part of your proof but Im guessing everything starts where it says "By the outer regularity"? and the first paragraph was just extra info? – alpastor May 22 '16 at 21:42
  • Yes, @NickE. The first part establishes the need for a further assumption, that $E$ itself is measurable. – Daniel Fischer May 22 '16 at 21:44
  • @DanielFischer Sorry, but I just want to make sure I understand you. Are you saying that in order to get the desired $G_\delta$ set, we need not assume that $E$ is measurable; but in order to get both the desired $G_\delta$ set and $F_\sigma$, we do need to assume that $E$ is measurable? – user193319 Feb 07 '19 at 00:39
  • @DanielFischer Also, to get just the $G_\delta$ set, do we even need to assume that $m^*(E) < \infty$? – user193319 Feb 07 '19 at 00:55
  • @user193319 Yes, for every $E$ there is a $G_{\delta}$-set $G$ with $E \subset G$ and $m^{\ast}[E] = m^{\ast}[G]$, regardless of whether $E$ is measurable or not, and of whether the outer measure is finite or not. For $m^{\ast}[E] < +\infty$, the argument in the question shows it. For infinite outer measure it's trivial, just take $G = \mathbb{R}$. It becomes more interesting then if you want a $G_{\delta}$ such that $m^{\ast}[E \cap I] = m^{\ast}[G \cap I]$ for every interval $I$. – Daniel Fischer Oct 02 '19 at 22:39