I am trying to understand the following proof of Lemma 25.3 in the book measure theory from Bauer. The Lemma states:

If $E$ is a Hausdorff space in which any point assets a countable neighbourhood basis, then any inner regular borel measure is locally finite.

The author requires for the borel measure that for every compact set C it follows $\mu(C) < \infty$. This is the first part of the proof:

The proof goes by contradiction. It is assumed that $\mu$ is a inner regular measure which is not locally finite. Then there exists a $x\in E$ such that for any open neighbourhood $V$ we have $\mu(V) =\infty$. From the requirement of the lemma, we find a sequence $V_1,V_2,\dots$ of open neighbourhoods, where we can assume that $$V_1 \supseteq V_2 \supseteq \dots$$ and $$ \bigcap_{i \in \mathbb{N}} V_i = \{x\}. $$

Why can I assume that $$ \bigcap_{i \in \mathbb{N}} V_i = \{x\} $$ holds? For the moment I only see that $$ \bigcap_{i \in \mathbb{N}} V_i = A $$ holds for some $A \subset E$ with $x \in A$. Anyway, even if I accept that I still struggle with the further proof.

Since $\mu(V_n) = \infty$ and the fact that $\mu$ is a inner regular measure we find for any $n \in \mathbb{N}$ a compact set $K_n \subseteq V_n$ with $\mu(K_n) > n$. Now the set $$K := \{x \} \cup \bigcup_{n=1}^\infty K_n$$ is compact, because for any open cover $(U_i)_{i\in I}$ of $K$ there exists a $i_0\in I$ with $x \in U_{i_0}$ and a $n_0 \in \mathbb{N}$ with $V_{n_0} \subseteq U_{i_0}$

By definition, we find a open neighbourhood $\tilde{V}$ in our neighbourhood basis of $x$ with $\tilde{V}\subseteq U_{i_0}$. But how can we guarantee that this $\tilde{V}$ is in our sequence? Of course one could force that we put the whole neighbourhood basis in our sequence, but then I dont see how one can guarantee to find a sorting of the basis where $$V_1 \supseteq V_2 \supseteq \dots$$ holds. The rest of the proof makes sense to me.

Furthermore, we have $K_n \subset U_{i_0}$ for all $n \geq n_0$. Therefore, $K$ can be covered by finitely many $U_i$'s. Because of $K_n \subseteq K$ it follows $$ n < \mu(K_n) \leq \mu(K)$$ and, therefore, $\mu(K) = \infty$. Hence, $\mu$ is no Borel measure which contradicts the assumption.

Would be great if someone could explain these two steps to me.


I’ll deal with the questions in order.

  1. The point $x$ has a countable nbhd base, say $\mathscr{B}=\{B_n:n\in\Bbb N\}$. $E$ is Hausdorff, so if $y$ is any other point of $E$, there is an $n\in\Bbb N$ such that $y\notin B_n$. Thus, $\bigcap_{n\in\Bbb N}B_n=\{x\}$. For $n\in\Bbb N$ let $V_n=\bigcap_{k\le n}B_k$; then $V_{n+1}\subseteq V_n$ for each $n\in\Bbb N$, and $\bigcap_{n\in N}V_n=\bigcap_{n\in\Bbb N}B_n=\{x\}$.

  2. Let $\mathscr{U}=\{U_i:i\in I\}$ be an open cover of $K$, where $I$ is some index set. The point $x$ is in $K$, so $x$ must belong to some member of $\mathscr{U}$; say $x\in U_{i_0}$, where $i_0\in I$. $\mathscr{B}$ is a nbhd base at $x$, so by definition there is some $n_0\in\Bbb N$ such that $x\in B_{n_0}\subseteq U_{i_0}$. Recall that $V_{n_0}=\bigcap_{k\le n_0}B_k$; thus, $x\in V_{n_0}\subseteq B_{n_0}\subseteq U_{i_0}$.

Note that whenever you have a countable nbhd base at a point, you can use this technique to get one that is nested; thus, you might as well assume from the start that your countable nbhd base is nested. (Warning: You cannot assume that an uncountable nbhd base is nested.)

  • $\begingroup$ To 1. I guess to obtain that $V_i$ is a open nbhd one needs to mention that for each $B_i$ there exists a open set $O_i \subseteq B_i$ and with the axiom of choice one can define $V_n := \bigcap_{k\leq n } O_k$ right? Thanks anyway, helped me a lot! $\endgroup$ – Adam Nov 14 '14 at 20:51
  • $\begingroup$ @Adam: You’re welcome! You don’t need anything extra if you take $\mathscr{B}$ to be a local base in the usual sense, i.e., of open nbhds: there’s no choosing to be done. $\endgroup$ – Brian M. Scott Nov 15 '14 at 15:51

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