Every tight $\tau$-additive finite measure is Radon

Question

According to the 7.2.2 Theorem of the book "Measure Theory" written by V.I. Bogachev, every tight $\tau$-additive finite measure is Radon. The proof says: "The restrictions of a $\tau$-additive measure to all compact subspaces are Radon". That theorem also says that a $\tau$-additive finite measure on a compact Hausdorff space is Radon, so I tried to use this fact. However, I couldn't prove that the restriction of a $\tau$-additive measure to a compact subspace is also $\tau$-additive.

Below I'll recall some definitions:

• A Borel finite measure $\mu$ on a topological space $X$ is called Radon if for all $B$ in $\mathcal{B}(X)$ and $\varepsilon >0$, there exists a compact set $K\subseteq B$ such that $\mu (B\setminus K)<\varepsilon$.
• A Borel measure $\mu$ on a topological space $X$ is called tight if for all $\varepsilon >0$ there exists a compact $K\subseteq X$ such that $\mu (X\setminus K)<\varepsilon $.
• A Borel measure $\mu$ on a topological space $X$ is called $\tau$-additive if for every increasing net of open sets $(U_\lambda)_{\lambda\in\Lambda}$ in $X$, one has the equality $\mu (\cup _{\lambda\in\Lambda}U_\lambda )=\lim_{\lambda}\mu (U_\lambda)$.

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My question is: how can I prove that every tight $\tau$-additive finite measure on a Hausdorff space is also Radon?

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I know that given a closed set $C\subseteq X$ and $\tau$-additive measure $\mu :\mathfrak{B}(X)\to \mathbb{R}$, the measure $\mu _C:\mathfrak{B}(X)\to \mathbb{R}$ given by $\mu _C(E):=\mu (E\cap C)$ is $\tau$-additive w.r.t. topology of $X$. However, I don't know if the restriction $\mu|_{\mathfrak{B}(F)}$ is $\tau$-additive w.r.t. subspace topology of $F$.

Answer 1

It seems that the part you're having trouble with is proving that the restriction of a bounded $\tau$-additive Borel measure to a Borel subset is $\tau$-additive (and you can follow the rest of Bogachev's proof once you know this). If you want to give it one last try, do so before crossing the following line.

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Let $X$ be a topological space, $\mu$ a bounded $\tau$-additive Borel measure on $X$, $S$ a Borel subset of $X$ and define $\nu$ to be the restriction of $\mu$ to $S$. Suppose for a contradiction that $\nu$ is not $\tau$-additive, so there exists a directed family $\mathcal{U}$ of $S$-open sets such that: $$ \sup_{U \in \mathcal{U}} \nu(U) < \nu\left(\bigcup \mathcal{U}\right). $$ For convenience, we write $U' = \bigcup \mathcal{U}$, and observe that the above implies there exists $\epsilon > 0$ such that for all $U \in \mathcal{U}$, $\nu(U' \setminus U) = \nu(U') - \nu(U) \geq \epsilon$.

To relate this back to the $\tau$-additivity of $\mu$, define $$ \mathcal{V} = \{ V \subseteq X \mid V \text{ open and } V \cap S \in \mathcal{U} \} $$ i.e. the set of open subsets of $X$ that restrict to elements of $\mathcal{U}$, and define $V' = \bigcup \mathcal{V}$. It is easily proved that $\mathcal{V}$ is directed and $V' \cap S = U'$. By $\tau$-additivity, there exists $V \in \mathcal{V}$ such that $\mu(V') - \mu(V) < \frac{\epsilon}{2}$.

Now, intuitively, we have reached a contradiction because there's a gap of measure at least $\epsilon$ between $U = V \cap S$ and $U' = V' \cap S$, but the gap between $V'$ and $V$ is less than $\frac{\epsilon}{2}$. Or in full algebra: $$ \frac{\epsilon}{2} > \mu(V' \setminus V) = \mu((V' \setminus V) \cap S) + \mu((V' \setminus V) \setminus S) \geq \mu((V' \setminus V) \cap S) = \nu(U' \setminus U) \geq \epsilon $$

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Here is some information from Fremlin's textbook Measure Theory to give you a more general overview of what is possible with $\tau$-additive measures and which assumptions were really necessary in the above proof.

The restriction of a finite $\tau$-additive Borel measure remains $\tau$-additive when restricted to an arbitrary subset $S$, i.e. it is not necessary for $S$ to be measurable - 414K

The above proof fails for $\tau$-additive finite measures defined on $\sigma$-algebras that are coarser than the Borel $\sigma$-algebra (i.e. that don't measure all open sets) - 414Y(c)

The above proof fails for infinite $\sigma$-finite measures. Fremlin has constructed a locally compact Hausdorff space $X$, a locally finite $\tau$-additive $\sigma$-finite Borel measure $\mu$, and a closed set $C \subseteq X$ with $\mu(C) = 1$ such that $\mu|_C$ is not $\tau$-additive - 419A (Fremlin does not define $\mu$ on the Borel $\sigma$-algebra, but on a $\sigma$-algebra containing all the open sets, and the above properties carry over to the restriction of $\mu$ to the Borel $\sigma$-algebra.)