Where completeness and separability are used in a proof of this lemma?

Question

I'm reading Lemma 2.30. from this note.

Lemma 2.30. Let $\mathcal{V}$ and $\mathcal{D}$ be two algebras of subsets of a separable complete metric space $T$. Assume $\mathcal{V} \subseteq \mathcal{D}$ and $\mu: \mathcal{D} \rightarrow[0, \infty)$ is (finitely) additive and for every $B \in \mathcal{V}$ we have $$ \mu(B)=\sup \{\mu(K): K \subseteq B, K \in \mathcal{D}, K \text { compact }\} $$ then $\mu$ is $\sigma$-additive on $\mathcal{V}$.

Below is the proof by the author.

Proof. Suppose not: there are $B_1, B_2, \ldots$ in $\mathcal{V}$ disjoint and $B \in \mathcal{V}$ such that $B=\bigcup_{i=1}^{\infty} B_i$ and $\delta:=\mu(B)-\sum_{i=1}^{\infty} \mu\left(B_i\right)>0$. Let $C_k:=B \backslash \bigcup_{j=1}^k B_j$. Then $C_k \in \mathcal{V}, C_k \supseteq C_{k+1}$, $\bigcap_{k=1}^{\infty} C_k=\emptyset$, and $\mu\left(C_k\right) \geq \delta$ for all $k$. Take for each $k$ a compact set $K_k \in \mathcal{D}$ with $K_k \subseteq C_k$ such that $\mu\left(C_k \backslash K_k\right)<2^{-k} \delta / 2$. Then $$ \begin{aligned} \mu\left(K_1 \cap \cdots \cap K_n\right) &=\mu\left(\left(C_1 \cap \cdots \cap C_n\right) \backslash \bigcup_{k=1}^n\left(C_k \backslash K_k\right)\right) \\ & \geq \mu\left(C_n\right)-\sum_{k=1}^n \mu\left(C_k \backslash K_k\right) \\ & \geq \mu\left(C_n\right)-\delta / 2 \geq \delta / 2, \end{aligned} $$

so $K_1 \cap \cdots \cap K_n \neq \emptyset$ for all $n$. Since the sets $K_k$ are compact it follows that $\bigcap_{k=1}^{\infty} K_k \neq \emptyset$. This contradicts $C_k \supseteq K_k$ and $\bigcap_{k=1}^{\infty} C_k=\emptyset$.

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My understanding:

• Because $\mathcal{V}$ is an algebra and $B, B_1, \ldots, B_k \in \mathcal{V}$, we get $C_k:=B \backslash \bigcup_{j=1}^k B_j \in \mathcal{V}$.

• Because $\mathcal{D}$ is an algebra and $C_k, K_k \in \mathcal{D}$, we get $C_k \setminus K_k \in \mathcal{D}$.

• The fact that $\mu$ is tight on every set in $\mathcal{V}$ justifies the existence of $K_k$'s. By finite additivity of $\mu$, we have $$ \mu\left(\left(C_1 \cap \cdots \cap C_n\right) \backslash \bigcup_{k=1}^n\left(C_k \backslash K_k\right)\right) \geq \mu\left(C_n\right)-\sum_{k=1}^n \mu\left(C_k \backslash K_k\right). $$

• Because compact sets in a metric space are closed, every $K_k$ is closed. Then we apply the finite-intersection characterization of compactness to get $\bigcap_{k=1}^{\infty} K_k \neq \emptyset$.

My question: Where the assumption that $T$ is complete separable, is used? It seems to me it's enough that $T$ is Hausdorff.

Answer 1

In fact, it's enough that $T$ is Hausdorff. This lemma is taken from Dudley's Real Analysis and Probability, i.e.,

Now given any two algebras $\mathcal{V} \subset \mathcal{D}$ of subsets of a set $S$, where $(S, \mathcal{T})$ is a Hausdorff topological space, and a finite, nonnegative, finitely additive function $\mu$ defined on $\mathcal{D}$, it will be said that $\mu$ is regular on $\mathcal{V}$ for $\mathcal{D}$ iff for every $B \in \mathcal{V}$, $$ \mu(B)=\sup \{\mu(K): K \subset B, K \in \mathcal{D}, K \text { compact }\} . $$

Lemma 10.2.4 If $\mathcal{V} \subset \mathcal{D}$ are two algebras, $\mu$ is finite and finitely additive on $\mathcal{D}$, and $\mu$ is regular on $\mathcal{V}$ for $\mathcal{D}$, then $\mu$ is countably additive on $\mathcal{V}$.