I somehow got stuck on the following line in Oxtoby: "The class of Borel sets that have an $F_\sigma$ subset and a $G_\delta$ superset of equal measure is a $\sigma$-algebra that includes all closed sets." We work here with a Borel measure (=measure defined for all Borel sets) on a metric space.

I have problems with showing that

$$\mathcal S=\{A\subseteq X; (\exists F\subseteq A\subseteq G) \mu(G\setminus F)=0, F\text{ is }F_\sigma, G\text{ is }G_\delta\}$$

is a $\sigma$-algebra.

As Oxtoby devotes only one grammatical sentence to this claim, I guess it should be relatively easy and I am overlooking some more straightforward approach.

What I tried so far:

Complements: $F\subseteq A\subseteq G$ $\Rightarrow$ $X\setminus G\subseteq X\setminus A\subseteq X\setminus F$.

Finite intersections: Let $F_1\subseteq A\subseteq G_1$, $F_2\subseteq B\subseteq G_2$ and $\mu(G_1\setminus F_1)=\mu(G_2\setminus F_2)=0$. Then $F_1\cap F_2\subseteq A\cap B\subseteq G_1\cap G_2$ and

$$G_1\cap G_2\setminus F_1\cap F_2 = (G_1\cap G_2\setminus F_1)\cup (G_1\cap G_2\setminus F_2)\subseteq$$

$$\subseteq (G_1\setminus F_1)\cup(G_2\setminus F_2),$$

hence $\mu(G_1\cap G_2\setminus F_1\cap F_2)=0$. (Also, $G_1\cap G_2$ is $G_\delta$ and $F_1\cap F_2$ is $F_\sigma$.)

Using finite intersections and complements, we can show that $\mathcal S$ is closed under differences of pairs of sets.

Disjoint countable unions: Let $F_i\subseteq A_i \subseteq G_i$, where $A_i$'s are disjoint measurable sets.

Then also $A_i\subseteq G_i\setminus \bigcup\limits_{n\ne i}F_n=: G_i'$ and $G_i'$'s are disjoint $G_\delta$-sets.

Thus we get

$$\bigcup_{i=1}^\infty F_i \subseteq \bigcup_{i=1}^\infty A_i \subseteq \bigcup_{i=1}^\infty G'_i$$

and $\mu(\bigcup_{i=1}^\infty F_i)=\mu(\bigcup_{i=1}^\infty G'_i)$.

But I do not know whether disjoint union of $G_\delta$ sets is again a $G_\delta$-set.

EDIT 2: As shown by Theo in the answer bellow, it is not true in general.

Using differences and disjoint countable unions, I could get arbitrary countable unions.

I also tried to prove that $\mathcal S$ is closed under countable intersections directly from the definition $\mathcal S$, but in that proof I obtained a countable intersection of $F_\sigma$-sets, which is not necessarily $F_\sigma$.


Based on Theo's suggestion: There are several equivalent characterizations of $\mathcal S$:

$$\mathcal S=\{A\subseteq X; (\forall \varepsilon>0)(\exists V\subseteq A\subseteq U) \mu(U\setminus V)<\varepsilon, V\text{ is close}, U\text{ is open}\}$$

$$\mathcal S=\{A\subseteq X; (\forall \varepsilon>0)(\exists F\subseteq A\subseteq U) \mu(U\setminus F)<\varepsilon, F\text{ is }F_\sigma, U\text{ is open}\}$$

Now I can modify the above construction to work for any countable union. For a system $\{A_i; i\in\mathbb N\}$ of sets from $\mathcal S$ I choose $F_i\subseteq A_i \subseteq U_i$ such that $\mu(U_i\setminus F_i)<\varepsilon.2^{-i}$, $U_i$ is open, $F_i$ is $F_\sigma$. Let $U:=\bigcup_{i=1}^\infty U_i$ and $F:=\bigcup_{i=1}^\infty F_i$. Then $U$ is open, $F$ is $F_\sigma$ and

$$U\setminus F= (\bigcup_{i=1}^\infty U_i) \setminus (\bigcup_{i=1}^\infty F_i) \subseteq \bigcup_{i=1}^\infty (U_i\setminus F_i),$$

hence $\mu(U\setminus F)<\varepsilon$. We also have

$$F=\bigcup_{i=1}^\infty F_i \subseteq \bigcup_{i=1}^\infty A_i \subseteq \bigcup_{i=1}^\infty U_i=U,$$

which implies $\bigcup_{i=1}^\infty A_i\in \mathcal S$.

  • 3
    $\begingroup$ Wouldn't it be easier to show first that the well-approximated Borel sets are precisely the Borel sets $A$ with the property that for every $\varepsilon \gt 0$ you can find a closed subset $F \subset A$ and an open superset $G \supset A$ such that $\mu(G \smallsetminus F) \lt \varepsilon$? Do I miss something? $\endgroup$ – t.b. May 26 '11 at 10:16
  • $\begingroup$ @Theo: I tried to prove the claim using modification of your hint. (I hope I haven't overlooked something.) I am not sure, if this is approximately the way you had in mind when you wrote your comment. $\endgroup$ – Martin Sleziak May 26 '11 at 12:59
  • 2
    $\begingroup$ This looks good and is more or less the argument I had in mind. I would probably have taken the $F_i$ closed, but the way you phrase it is actually cleaner (so you don't need the first characterization you stated). For the record -- I guess that should be obvious to you now: you can of course play around even more with $G_{\delta}$ in place of open, etc. $\endgroup$ – t.b. May 26 '11 at 13:00
  • $\begingroup$ @Theo: Thanks again for your help. Could you please post your comment as an answer, so that it can be accepted. meta.math.stackexchange.com/questions/1148/… $\endgroup$ – Martin Sleziak May 27 '11 at 7:55

Instead of just re-posting my first comment as an answer, I thought it would be better to comment briefly on your original question. (As I said in my second comment, your solution in the edit works fine).

Unfortunately (or rather fortunately) your original approach doesn't work as you intended. Here are three simple observations:

  1. Every closed set in a metrizable space is a $G_{\delta}$. In particular every point is a $G_{\delta}$.
  2. Recall the following form of the Baire category theorem: "the intersection of countably many dense $G_{\delta}$s in a complete metric space is again a dense $G_{\delta}$". This immediately implies that a countable dense subset of a perfect Polish space (e.g. $\mathbb{Q}$ in $\mathbb{R}$) is an $F_{\sigma}$ that isn't a $G_{\delta}$.
  3. It is obvious from the second sentence in 1. that a countable dense set is a countable disjoint union of $G_{\delta}$'s. But as I argued, it won't be a $G_{\delta}$ itself in general.

Finally, let me just note that the detour I suggested in my first comment is not strictly necessary, as by unpacking the proof of the equivalent descriptions you should be able to see what to do in order to save your original direct approach. However, it seems rather obfuscating than clarifying what is going on, so I leave that to you in case you're interested.

  • 2
    $\begingroup$ Just do make sure that I understand your point 2 correctly: If $D=\{d_n; n\in\omega\}$ is a dense set in $X$, where $X$ is a complete metric space with no isolated points, then $X\setminus D=\bigcap_{n\in\omega} X\setminus\{d_n\}$ is an intersection of dense (since $d_n$ is not-isolated) open sets, thus it is $G_\delta$ dense. If $D$ would be a $G_\delta$ set, then $D\cap (X\setminus D)=\emptyset$ would be dense, too (as intersection of dense $G_\delta$ sets), which is a contradiction. $\endgroup$ – Martin Sleziak May 27 '11 at 14:46
  • $\begingroup$ @Martin: That's it exactly. $\endgroup$ – t.b. May 27 '11 at 14:59

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