# Are regular measures continuous?

I'm taking a course on Measure Theory, and we are building measures from the very beggining, starting with semi-algebras $$\mathcal{S}$$ and proving extension theorems to get to measures on $$\sigma$$-algebras.

Now, we have proved Caratheodory's Extension Theorem, asserting that we can extend a $$\sigma$$-additive measure defined on a semi-algebra $$\mathcal{S}$$ or on the algebra generated $$\mathcal{A}(\mathcal{S})$$ to a $$\sigma$$-additive measure on the $$\sigma$$-algebra $$\mathcal{F}(\mathcal{S})$$ (uniquely if we started with a $$\sigma$$-finite measure).

The plan now is to use this to construct the Lebesgue measure $$\lambda$$ on $$\mathbb{R}$$. So we must find a way to prove that the Lebesgue measure is $$\sigma$$-additive on the algebra os intervals $$\mathcal{A}(\mathcal{S})$$.

The proof presented in class was a particular instance of the a general fact: if $$\mu$$ is a finitely additive and regular measure defined on an algebra, then it is $$\sigma$$-additive.

However, I was wondering if it would be possible to take a different approach. This fact seems to be heavily dependent on the topological properties of the underlying space, but I was wondering if the (slightly) more general result is valid:

If $$\mu:\mathcal{A}\to[0,+\infty]$$ is a finitely additive and regular measure defined on an algebra $$\mathcal{A}$$, then it is continuous from below.

It is possible to prove that continuity from below implies sigma additive, so this is a slightly more general result.

This is my attempt at a proof:

Let $$E_k, E\in\mathcal{A}$$, where $$E_k$$ increases to $$E$$, i.e., $$E_k\subset E_{k+1}$$ and $$E = \cup E_k$$. For any $$\varepsilon>0$$, by regularity, there is a compact set $$K\subset E$$, $$K \in \mathcal{A}$$ such that

$$$$\mu(E) - \varepsilon < \mu(K) \leq \mu(E)$$$$

My plan is to show that, whatever is $$K$$, there is an $$n$$ such that $$\mu(K)\leq\mu(E_n)$$. This way, when we take the supremum over all compact sets $$K\subset E$$, we get that $$\mu(E_n)\to\mu(E)$$.

I have tried various approaches to prove this, but I have not been able to succeed.

Edit 1: As suggested, I'm stating the definition for regularity in this context.

A measure $$\mu:\mathcal{S}\to [0,+\infty]$$ defined on a class of sets $$\mathcal{S}$$ in a topological space is said to be regular if, for every $$A\in\mathcal{S}$$:

$$$$\mu(A) = \inf\{\mu(G) | A\subset G, G\in\mathcal{S}, G \text{ open}\} = \sup\{\mu(K) | K \subset A, K \in \mathcal{S}, K \text{ compact}\}$$$$

• We need to be a little more careful with definitions: "regular" makes sense for a measure on the Borel $\sigma$-algebra of a topological space (though different authors use different definitions so it's a good idea to state your definition), but not for a measure on a general algebra of sets because the latter has no concept of "open", "compact", etc. – Nate Eldredge Aug 31 '20 at 21:04
• Are you aware that $\sigma$-additivity also implies continuity from below? So your "more general" result is actually not any more general. – Nate Eldredge Aug 31 '20 at 21:07
• Yes, I am aware of that! I'll edit to state my definition of regularity, that's a good idea. – André Muchon Aug 31 '20 at 21:13

Take $$A_n$$ sequence of increasing sets of $$\mathcal{A}$$ such that $$A = \underset{n \geq 1}{\bigcup}A_n \in \mathcal{A}$$. We wish to show that $$\lim \mu(A_n) = \mu(A)$$
Since $$A_n \subset A, \forall n \geq 1$$, then $$\mu(A_n) \leq \mu(A) \implies \lim \mu(A_n) \leq \mu(A)$$. For the other inequality, we may view this increasing sequence of sets as an increasing disjoint union of sets by defining $$B_1 = A_1$$ and $$B_n = A_n-A_{n-1}, \ n \geq 2$$. We have $$B_n \in \mathcal{A}, \ \forall n \geq 1$$ since $$\mathcal{A}$$ is an algebra and $$A_m = \sum\limits_{n=1}^mB_n$$ where the sum notation is used to denote a union of pairwise disjoint sets. Notice that since $$\mu$$ is finitely additive, then $$\mu(A_m) = \sum\limits_{n=1}^m\mu(B_n)$$ and now we will use that the measure is regular:
Let $$\epsilon>0$$ and $$U_n$$ be an open set such that $$B_n \subset U_n$$ and $$\mu(B_n) \leq \mu(U_n) \leq \mu(B_n) + \frac{\epsilon}{2^n}$$ and let $$K \subset A$$ be a compact set. Notice that $$K \subset A = \sum\limits_{n\geq 1}B_n \subset \underset{n\geq 1}{\bigcup}U_n$$, therefore by compactness of $$K$$, there exists $$m'$$ such that $$K \subset \bigcup\limits_{n=1}^{m'}U_n$$ which implies $$\mu(K) \leq \mu(\bigcup\limits_{n=1}^{m'}U_n) \leq \sum\limits_{n=1}^{m'}\mu(U_n) \leq \sum\limits_{n\geq 1}\mu(B_n) + \epsilon$$, since $$\epsilon$$ is arbitrary, then $$\mu(K) \leq \sum\limits_{n\geq 1}\mu(B_n) = \underset{m \to \infty}{\lim}\sum\limits_{n=1}^m\mu(B_n) = \underset{m \to \infty}{\lim}\mu(A_m)$$, since this is true for every compact set contained in $$A$$, and $$\mu(A)$$ is the supremum of the measures of all such compact sets, then $$\mu(A) \leq \lim\mu(A_n)$$.