4
$\begingroup$

Let $(X,\mathcal{T})$ be a topological space and $\mu$ be a content on a semiring $\mathfrak{J}$ over $X$. Suppose that $\mu$ is inner regular, i.e., for every $\epsilon > 0$, $A \in \mathfrak{J}$, there is a $K\in \mathfrak{J}$ such that $\overline{K}$ is compact, $\overline{K} \subseteq A$ and $\mu(A)\leq \mu(K)+\epsilon$. I want to show that this implies that $\mu$ is a premeasure.


My attempt:

First of all, I showed that if $\mu$ is an inner regular content on $\mathfrak{J}$, then its extension to the ring $\mathfrak{R}$ (which I denote by $\mu$ as well) generated by $\mathfrak{J}$ is inner regular as well. This allows us to use the following lemma:

Lemma: Let $\mu$ be a content on a ring $\mathfrak{R}$. If for every sequence $(A_n)_{n\in\mathbb{N}}$ of sets in $\mathfrak{R}$ satisfying $\mu(A_1) < \infty$ and $A_n \downarrow \emptyset$ we have $\mu(A_n)\downarrow 0$, then $\mu$ is a premeasure on $\mathfrak{R}$.

So, in order to use the lemma, let $A_n$ be defined as in the lemma. Take $K_n \in \mathfrak{R}$ such that $\overline{K_n}$ is compact, $\overline{K_n} \subseteq A_n$ and $\mu(A_n) \leq \mu(K_n) + \epsilon$ for all $n \in \mathbb{N}$. We have $$\bigcap_{n=1}^{\infty} \overline{K_n} \subseteq \bigcap_{n=1}^{\infty} A_n = \emptyset.$$ Due to the compactness of the $\overline{K_n}$, this means that the family $(\overline{K_n})_{n\in\mathbb{N}}$ does not have the finite intersection property. Hence, there is a $N \in \mathbb{N}$ such that $$ \bigcap_{n=1}^{N} K_n \subseteq \bigcap_{n=1}^{N} \overline{K_n} = \emptyset.$$ I've tried to use $\cap_{n=1}^{N} A_{n} = A_{N}$ and estimate $\mu(\cap_{n=1}^{N} A_n)$ by something depending on $\mu(\cap_{n=1}^{N} K_n)$ plus something depending on epsilon, but without success. I've also tried to make the $K_n$ disjoint, that is, define $C_n := K_n \setminus \cup_{i=1}^{n-1}K_i$ and then use that

$$\mu(A_1) \geq \mu\left(\bigcup_{n=1}^{N} C_n\right) = \sum_{n=1}^{N} \mu(C_n).$$

As $N$ was arbitrary, this shows that $\mu(C_n) \to 0$, but I've been unable to conclude from this.

$\endgroup$
4
+100
$\begingroup$

Let $A_n$ be defined as in the lemma. Fix $\epsilon>0$ and choose $K_n \in \mathfrak{R}$, $\overline{K_n}$ compact and $\overline{K_n} \subseteq A_n$, such that $\mu(A_n)<\mu(K_n) + \epsilon/2^n$. By what you've already shown, there's an $N$ such that $\bigcap_{n=1}^N K_n = \emptyset$. Now,

\begin{align} \mu(A_N) &= \mu(\cap_{n=1}^N A_n) \leq \mu(\cup_{n=1}^N (A_n - K_n))\\ &\leq \sum_{n=1}^N \mu(A_n - K_n) = \sum_{n=1}^N \big( \mu(A_n) - \mu(K_n)\big) < \epsilon. \end{align}

Since $(A_n)$ is decreasing, $\mu(A_n) < \epsilon$ for all $n \geq N$, so $\mu(A_n) \to 0$.

$\endgroup$
  • 1
    $\begingroup$ Nice! It took me a little to get that we have $$\bigcap_{n=1}^{N} A_{n} = \left(\bigcap_{n=1}^{N} A_{n}\right) \setminus \left(\bigcap_{n=1}^{N} K_n\right) \subseteq \bigcup_{n=1}^{N}A_{n} \setminus K_{n}$$ to understand where the first inequality was coming from. $\endgroup$ – user159517 Jul 29 '18 at 22:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.