# Inner regular content is premeasure

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.

## 1 Answer

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 $\mu(A_n) \to 0$.

• 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. – user159517 Jul 29 '18 at 22:36