Checking measurability on open sets

Question

This is exercise 5 of section 53 in Halmos' Measure theory.

Let $X$ be a locally compact Hausdorff space and $\mu^{*}$ an outer measure on the hereditary class of $\sigma$-bounded sets. Suppose $\mu^{*}(C)=\inf_{C \subset U, U \: \text{open}} \mu^{*}(U)<+\infty$ for every compact $C$.

Let $E$ be a $\sigma$-bounded set such that \begin{equation*} \mu^{*}(U)=\mu^{*}(U\cap E) + \mu^{*}(U\cap E^{c}) \end{equation*} for every open $U$. Is it true that $E$ is $\mu^{*}$-measurable ?

My guess is no, but I can't find a counter example.

Answer 1

$E$ is $\mu* measurable $

$\mu^{*}(U)=\mu^{*}(U\cap E) + \mu^{*}(U\cap E^{c})$ is also known as the Carathéodory's Criterion see here Any set $E$ satisfying this criterion is measurable. Remember $U$ is not even required to be measurable.

Answer 2

An example.

Update: Unfortunately, this is not a counterexample. The outer measure is not finite on compacts. The example can be modified (by removal of the point $\{(x^*,y^*)\}$) to ensure the finiteness on compacts but then the immeasurable set in question would not be $\sigma$-finite.

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Consider two uncountable sets $X$, $Y$, both with the discrete topologies, and their one point compactifications: $X^* = X \cup \{x^*\}$ and $Y^* = Y \cup \{y^*\}$. Let also $A\subset X$ be such that both $A$ and $X\setminus A$ are uncountable.

Now, $$ Z = X^* \times Y^* $$ with the product topology is a compact Hausdorff space. Define an outer measure $\phi$ on $\mathcal P(Z)$: $$ \phi(E) := \sum_{y\in Y^*} \mu_0(E_y) + \infty \cdot \chi_E(x^*,y^*), $$ where $E_y := \{x\in X^*: (x,y) \in E\}$ and $$ \mu_0(F) = \begin{cases} 1, &\text{if $F$ is uncountable or } x^*\in F,\\ 0, &\text{otherwise,} \end{cases} \qquad \text{ for } F\subset X^*. $$ The infinity term in the definition of $\phi$ (meaning that $\phi(\{(x^*,y^*)\}) = +\infty$) is to ensure outer regularity of the singleton $\{(x^*,y^*)\}$. Similarly, I choose $\mu_0$ so that $\mu_0(\{x^*\}) = 1$, which guarantees outer regularity of $\{x^*\}$. This will be seen later.

Clearly, the set $A\times \{y^*\}$ is not $\phi$-measureable because $$ 1 = \phi(X\times \{y^*\}) < \phi(A\times \{y^*\}) + \phi((X\setminus A)\times \{y^*\}) = 2. $$

However, for any open $U\subset Z$: $$ \phi(U) \ge \phi(U\setminus A\times\{y^*\}) + \phi(U\cap A\times\{y^*\}). $$ To see this, take $U$ such that $\phi (U\cap A\times\{y^*\})>0$, which means that $U\cap A\times\{y^*\}$ is uncountable. And if $(x,y^*) \in U$, then necessarily $\{x\}\times Y^* \cap U$ is cofinite. Then $U_y$ is uncountable for infinitely $y$'s, and consequently, $\phi(U) = +\infty$.

The only thing left is the outer regularity of $\phi$ on compacts. Take a compact $K$ of finite measure and consider compacts $K_y$ for $y\in Y^*$. I'll show that for any such compact, there exists an open set $U(y) \supset K_y\times \{y\}$ of the same measure.

Case $y\neq y^*$. If $K_y$ doesn't contain $x^*$, then it's already open (and finite). On the other hand, if $x^*\in K_y$ then $\mu_0(K_y) = 1$ and $K_y\times \{y\}$ can be approximated by $X^* \times \{y\}$.

Now, consider $K_{y^*}$. It doesn't contain $x^*$ because $K$ has finite measure. So, it must be finite (and of measure 0). An open set that approximates it is $K_{y^*} \times Y^*$, which also has zero measure.

Answer 3

I'm going to answer my own question. The answer is no, it is not enough to check measurability on open sets.

Let $A,B,C$ be three sets that partition $[0,1]$, that have Lebesgue inner measure $0$ and Lebesgue outer measure $1$ (see section 16 in Halmos' book). Write $\lambda$ for Lebesgue measure.

Knowledge of $X\cap A$, with $X$ a Borel set, determines $X$ up to a $\lambda$-negligible set: if $(X\Delta Y)\cap A = \emptyset$ then $[0,1]-X\Delta Y$ covers $A$ so that $\lambda(X\Delta Y)=0$ because $A$ has outer measure $1$.

The sets of the form $(X\cap A) \cup (Y \cap B)\cup (Z\cap C)$ with Borel $X,Y,Z$ constitue a $\sigma$-algebra $\mathcal{A}$. For every $E\in \mathcal{A}$ define \begin{align*} \mu(E)&=\frac{\lambda(Y)}{2} \: \text{if} \: \lambda(X)=\lambda(Z)=0, \\ &= \frac{\lambda(X)+\lambda(Y)+\lambda(Z)}{3} \: \text{otherwise}. \end{align*}.

It is easy to check that $\mu$ is monotone and countably subadditive, so it can be extended to an outer measure on $\mathcal{P}([0,1])$ by setting $\mu(E)=\inf_{F\in \mathcal{A}}\mu(A)$. Also for every compact set $C$ \begin{equation*} \mu(C)=\lambda(C)=\inf_{C\subset U} \mu(U) \leqslant 1 \end{equation*} and for every open set $U$ \begin{equation*} \mu(U)=\mu(U\cap A) + \mu(U \cap A'). \end{equation*} Yet \begin{equation*} 2/3=\mu(A\cup B) < \mu(A) + \mu(B) = 5/6 \end{equation*} so that $A$ is not measurable.