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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

\begin{equation}\mu(E) - \varepsilon < \mu(K) \leq \mu(E) \end{equation}

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 $A\in\mathcal{S}$:

\begin{equation}\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}\} \end{equation}

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  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge Aug 31 '20 at 21:04
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    $\begingroup$ Are you aware that $\sigma$-additivity also implies continuity from below? So your "more general" result is actually not any more general. $\endgroup$ – Nate Eldredge Aug 31 '20 at 21:07
  • $\begingroup$ Yes, I am aware of that! I'll edit to state my definition of regularity, that's a good idea. $\endgroup$ – André Muchon Aug 31 '20 at 21:13
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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 $\mu(A) \leq \lim\mu(A_n)$.

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