# If the finitely additive measure of an open set is approximable by clopen sets, is it approximable from within?

Let $$F$$ be a finite set equipped the discrete topology. Let $$X = F \times F \times ...$$ be the countably infinite product space equipped with the product topology. Let $$\mathcal A$$ be any field of subsets of $$X$$ that contains the open subsets of $$X$$. Let $$\mu$$ be a finitely additive, finite measure with domain $$\mathcal A$$.

Suppose that $$\mu$$ has the following "clopen approximation property": For any $$\epsilon > 0$$ and any open subset $$G$$ of $$X$$, there is a clopen subset $$C$$ of $$X$$ such that $$\mu(G \triangle C)< \epsilon$$.

Is the clopen approximation property equivalent to the following "inner regularity property"? For every open subset $$G$$ of $$X$$, $$\mu(G) = \sup\{\mu(C): C \subseteq G, C \text{ clopen}\}$$.

Clearly inner regularity implies clopen approximation, but I am unable to see that the converse is true.

If $$G$$ is open, then it can be written as a countable union $$G = C_1 \cup C_2 \cup ...$$ of pairwise disjoint clopen sets. It seems reasonable to expect that if $$G$$ is approximable by some sequence of clopen sets, then it should be approximable from within by finite unions of its constiuent clopen sets. i.e. $$\mu(\cup_{i=1}^n C_i) \to \mu(G)$$ as $$n \to \infty$$. But, again, I am unable to see that this is the case.

If $$G$$ is approximable by some clopen sequence $$B_n$$, so that $$\mu(G \triangle B_n) \to 0$$, then $$\mu(G - B_n) \to 0$$ and $$G - B_n$$ is an open subset of $$G$$. If we could replace this sequence of open subsets of $$G$$ with a "similar" sequence of clopen subsets of $$G$$, maybe we could prove the result. I played around with sequences like $$(C_1 \cup ... \cup C_n) - B_n$$ (recall that $$G = C_1 \cup C_2 \cup ...$$), but didn't get anywhere.

The motivation for this has to do with finding countably additive extensions of finitely additive measures. The clopen approximation property can be used to characterize extreme points in the convex set of extensions from the clopen field to another, larger field. And, since the clopen sets form a compact class, inner regularity provides a sufficient condition for countable additivity. If the result in question holds, then I could say that the extreme points of the set of extensions from the clopen field to the field generated by open sets are countably additive on the latter.

• What do you mean more precisely by $X=F\times F \times \ldots$? A finite or infinite product? – Skeeve Mar 20 at 16:00
• @Skeeve Infinite. This is standard notation, right? A finite product would be $F \times ... \times F$. Anyway, I will edit and clarify this, thanks. – aduh Mar 20 at 16:02

The answer to this question is negative. To construct a counterexample, fix any free ultrafilter $$\mathcal U$$ on $$X$$ containing a discrete subspace $$D$$ of $$X$$. The characteristic function $$u:\mathcal P(X)\to \{0,1\}$$ of this ultrafilter determines a $$\{0,1\}$$-valued finitely additive measure $$u$$ defined on the algebra $$\mathcal P(X)$$ of all subsets of $$X$$.

We claim that for every subset $$S\subset X$$ there exists a clopen set $$C\subset X$$ with $$u(C\triangle S)=0$$. Indeed, if $$u(S)=0$$, then for the clopen set $$C=\emptyset$$ we have $$u(C\triangle S)=u(S)=0$$. In $$u(S)=1$$, then for the clopen set $$C=X$$ we have $$u(C\triangle S)=u(X\setminus S)=u(X)-u(S)=1-1=0$$.

Let $$A$$ be the (closed) set of accumulation points of the discrete subspace $$D$$ in the compact metrizable space $$X$$. The open set $$U=X\setminus A$$ contains the set $$D$$ and hence has measure $$u(U)=1$$. On the other hand, for any clopen set $$K\subset U$$, the set $$K$$ is compact and disjoint with $$A$$. Then the intersection $$D\cap K$$ is finite and hence $$K\notin \mathcal U$$, which means that $$u(K)=0$$. Therefore, $$u(U)=1\ne 0=\sup\{u(K):K\subset U\mbox{ is a clopen subset of X}\}.$$

Now we shall modify the measure $$u$$ in order to make it strictly positive.

Let $$p$$ be the standard product probability measure defined on the $$\sigma$$-algebra $$\mathcal B(X)$$ of Borel subsets of $$X$$. It is clear that $$p$$ is strictly positive in the sense that $$p(U)>0$$ for any non-empty open subset $$U\subset X$$.

Consider the strictly positive probability measure $$\mu=\frac12u+\frac12p$$, defined on the $$\sigma$$-algebra $$\mathcal B(X)$$.

We can chose the discrete subset $$D$$ so that its closure $$\bar D=A\cup D$$ has product measure $$p(\bar D)=0$$. Then the open set $$X\setminus A$$ has measure $$\mu(X\setminus A)=1\ge \frac{1}{2}=\sup\{\mu(U):U\subset X\setminus A\mbox{ is a clopen set in X}\}.$$

Now we prove that for any Borel subset $$B\subset X$$ and every $$\varepsilon>0$$ there exists a clopen subset $$C\subset X$$ such that $$\mu(B\triangle C)<\varepsilon$$.

If $$u(B)=0$$, then by the regularity of the measure $$p$$, find a compact subset $$K\subset B\setminus A$$ such that $$p(K)>p(B)-\varepsilon$$. Using the regularity of the measure $$p$$ once more, find a clopen subset $$C\subset X$$ such that $$K\subset C\subset X\setminus A$$ and $$p(C). Then $$p(C\triangle B)\le p(C\triangle K)+p(K\triangle B)<\varepsilon+\varepsilon=2\varepsilon$$ and $$\mu(C\triangle B)=\tfrac12p(C\triangle B)+\tfrac12u(C\triangle B)<\tfrac122\varepsilon+\tfrac12u(C)+\tfrac12u(B)=\varepsilon+0+0=\varepsilon.$$

If $$u(B)=1$$, then by the regularity of the measure $$p$$, there exists a compact subset $$K\subset B$$ such that $$A\subset K$$ and $$p(K)>p(B)-\varepsilon$$. Using the regularity of the measure $$p$$ once more, find a clopen subset $$C\subset X$$ such that $$K\subset C$$ and $$p(C). Then $$p(C\triangle B)\le p(C\triangle K)+p(K\triangle B)<\varepsilon+\varepsilon=2\varepsilon.$$ Since $$C$$ is a neighborhood of $$A$$, the set $$D\setminus C$$ is finite and hence $$u(C\cap D)=u(D)-u(D\setminus C)=1-0=1$$. Since $$\mathcal U$$ is a filter, $$C\cap B=1$$ and hence $$u(C\cap B)=1$$ and $$u(C\triangle B)=0$$. Then $$\mu(C\triangle B)=\tfrac12p(C\triangle B)+\tfrac12u(C\triangle B)<\tfrac122\varepsilon+0=\varepsilon.$$

• Thanks very much. One follow up question: How about if we require that the $\mu$ measure of every clopen set be non-zero? Is the answer to the question affirmative in that case? – aduh Mar 20 at 19:26
• It seems that a convex combination of the ultrafilter measure and the product measure will have the property you require: the measure of any non-empty clopen set is strictly positive. – Taras Banakh Mar 20 at 19:34
• But for such $\mu$, the argument in the second paragraph no longer works. Or am I missing something? – aduh Mar 20 at 19:55
• @aduh Yes, this argument should be modified a little bit. We can choose $D$ so that $A$ has measure zero in the product measure. This will help to choose a suitable clopen set $C$. – Taras Banakh Mar 20 at 20:25
• I don't think I see the idea yet--I guess I'm a bit slow. If you could spell out the details, that would be helpful. Thanks again. – aduh Mar 20 at 20:53