

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

I'm reading this book Probability & Measure Theory by Ash. I think I've come across a part that is a little hand-wavy. We are trying to build a Lebesgue-Stieltjes measure from a distribution function $F$ (in that the measure of interval $(a,b]$ is $F(b) - F(a)$).

He starts by adding $+\infty$ and $-\infty$ to the real line so that we can work in compact space. He defines right-semiclosed as intervals of the form $(a, b]$ and $[-\infty, b]$ and $(-\infty, b]$. He then constructs a field by taking all finite unions of these right-semiclosed intervals.

He defines a set function over this field defined in the intuitive way (the set function takes $(a,b]$ to $F(b) - F(a)$), and he shows that this set function is countably additive.

This is where I don't understand his argument. He seems to say, ignore these points $+\infty$ and $-\infty$ so that our field no longer uses the compact space, and our set function now becomes a proper measure over a real field. Then apply the Carathéodory Extension Theorem.

I don't see how we can go from a compact space to a non-compact space without causing harm to the properties of our set function. I'm hoping that this construction method is widely used, and someone can explain where I am confused. This is Theorem 1.4.4 in Ash, 2nd Edition.

The complete exposition can be found at from the bottom of page 22 to page 24.

share|cite|improve this question
I don't have the book, so I don't know where does the need to work in a compact space arises. From what I know, Caratheodory's extension theorem works in this generality with no problem. – Mark Jul 4 '11 at 21:05
Working in a compact space allows Ash to show that his original set function is countably additive over his field. I have edited the post to include a link to the exposition. – Mark Jul 4 '11 at 21:16
One assume you mean $F(b)-F(a)$ instead of $F(a)-F(b)$. – GEdgar Jul 4 '11 at 21:35
@GEdgar - Yes, thanks for that catch. – Mark Jul 4 '11 at 22:26
I don't see where he "ignores" the two points added, so to speak. Maybe this is outside the Google preview. – Mark Jul 5 '11 at 0:26

There is an equivalence between Stieltjes measures on the real line, and Stieltjes measures on the compactified real line that assign measure 0 to the added boundary points at infinity.

For any measure of one type there is a unique measure of the other type that assigns the same values to finite intervals in $R$. This is nothing more than the notation of "improper integrals" from calculus. As in calculus it is a notational convenience used to avoid constantly writing about limits of integrals. The proof could be written or read in terms that avoid any extension of the space or the measure, just as any calculation with improper integrals can be presented as a limit of calculations on finite intervals.

share|cite|improve this answer
I'm not sure about your argument about notational convenience. The author explicitly uses a topological argument which needs the compactness of the closure of the reals. – Mark Jun 16 '13 at 16:16
The construction of Stieltjes measures is order-theoretic in nature, and I strongly suspect that any (sequential) compactness required is in the range of $F$, not its domain. The distribution function is real valued and has finite range, which is a closed interval upon extension to $[-\infty,+\infty]$. What else does the author say is needed? – zyx Jun 17 '13 at 2:27
I guess google removed the ability to look inside the book. The author uses the finite intersection property during the proof that the set function u(a,b] = F(b)-F(a) is countably additive. He makes a collection of closed sets limiting down to the empty set, and uses compactness to argue that it only takes a finite number of intersection to reach the empty set. I agree that you can probably prove countable additivity without compactifying the reals, but then I guess you have to some additional work to handle intervals whose unions are unbounded. – Mark Jun 18 '13 at 4:14
These arguments are (as far as I can imagine without ability to read the preview) essentially that $F$ commutes with sequential limits, after extension to $[-\infty, +\infty]$. But this is another way of restating the properties and purpose of the "improper integrals" notation. – zyx Jun 19 '13 at 16:07

I agree this does seem a little hand-wavy, and as far as I can tell it's completely unnecessary! From my experience we may rigorously construct Lebesgue measure on the real line as follows.

Let $\mathcal{D}$ be the set of all finite unions of intervals of the form $(a,b]$ for $a\leq b \in \mathbb{R}$. We allows $a=b$ so that $\mathcal{D}\supset\emptyset$. It is easily checked that $\mathcal{D}$ is a ring in the sense of measure theory. Also it's fairly obvious that $\mathcal{D}$ generates the Borel $\sigma$-algebra on $\mathbb{R}$.

We define a set function on $\mathcal{D}$ by $$\mu(\bigcup_{n=1}^k(a_n,b_n])=\sum_{n=1}^k (b_n-a_n)$$

This is well-defined, as you can check. It turns out that the hard bit to prove is countable additivity, where it looks like your book took a slightly unclear shortcut. I'll prove it here now slowly.

There's a well-known lemma which states that an additive set function is countably additive iff whenever $A_n$ a decreasing sequence of measurable sets with $\bigcap A_n=\emptyset$ we have $\mu(A_n)\rightarrow 0$. It's not too hard to prove and is in all the major texts.

We now use this lemma and argue by contradiction. Suppose we have such $A_n$ and $\exists \epsilon >0$ s.t. $\mu(A_n)>2\epsilon$ $\forall n\in\mathbb{N}$. Now choose some $C_n\in\mathcal{D}$ with $\bar{C_n}\subset A_n$ and $\mu(B_n\setminus C_n)<\epsilon 2^{-n}$. You can check that

$$\mu(B_n\setminus(C_1\cap\dots\cap C_n))<\epsilon$$

and hence that $\mu(C_1\cap\dots\cap C_n)>\epsilon$ for all $n$.

Thus in particular $K_n=\bar{C_1}\cap\dots\cap \bar{C_n}\neq\emptyset$. But the $K_n$ are a decreasing sequence of closed subsets of a complete space with diameter tending to zero, so by standard topology $\bigcap K_n\neq \emptyset$. But then certainly $\bigcap A_n\neq \emptyset$, a contradiction.

Now we've got countably additivity sorted, we can extend to the Borel $\sigma$-algebra by Caratheodory.

Summary: this method for proving countable additivity seems a little less esoteric to me. All the other details are essentially very similar to your book!

share|cite|improve this answer
Here the measure of an interval is not $b-a$, but $F(b)-F(a)$. – zyx May 27 '12 at 20:30
Ah damn misread the question. I wonder whether my approach can be fixed though... I'll have a think about it! – Edward Hughes May 27 '12 at 20:33
up vote 0 down vote accepted

zyx's answer has helped me understand the proof.

Let $F: \mathbb{R}\rightarrow\mathbb{R} $ be a distribution function. We can form the function $F_C:\overline{\mathbb{R}}\rightarrow\overline{\mathbb{R}} $ by setting $$F_C(x) \equiv F(x), x\in\mathbb{R}$$ $$F_C(\pm\infty) \equiv \lim_{x\rightarrow \pm\infty} F(x)$$

Then we can define set function $$\mu_C(a,b] \equiv F_C(b)-F_C(a), a\in\overline{\mathbb{R}}, b\in \overline{\mathbb{R}}$$ $$\mu_C[-\infty, b] \equiv \mu_C(-\infty, b], b \in \overline{\mathbb{R}}$$

We can prove $\mu_C$ is countably additive on the field of finite unions of disjoint right closed intervals in $\overline{\mathbb{R}}$.

Now we define a set function $\mu$ on the field of finite unions of disjoint right closed intervals in $\mathbb{R}$:
$$\mu(a,b] \equiv \mu_C(a,b], a\in\mathbb{R}, b\in \mathbb{R}$$ $$\mu(-\infty, b] \equiv \mu_C(-\infty, b], b \in \mathbb{R}$$ $$\mu(a, \infty) \equiv \mu_C(a, \infty], a \in \mathbb{R}$$ $$\mu(-\infty, \infty) \equiv \mu_C[\infty, \infty], a \in \mathbb{R}$$

Now suppose we have a sequence of disjoint sets $A_n$ in the field on $\mathbb{R}$, and $\cup A_n = A$ and $A$ is in the field on $\mathbb{R}$. $$\mu(A) = \mu(\cup A_n)$$ $$\mu(\cup A_n) = \mu(A_0 \cup A_1 \cup (\cup_{n =2} ^{\infty} A_n))$$ where $A_0$ and $A_1$ are renumbered to be the unbounded sets $(a, \infty)$ and $(-\infty, b)$ if they exist. Otherwise we can take them to be the empty set. $$\mu(A_0 \cup A_1 \cup (\cup_{n =2} ^{\infty} A_n)) = \mu(A_0) + \mu(A_1) + \mu (\cup_{n =2} ^{\infty} A_n)$$ by finite additivity. $$\mu(A_0) + \mu(A_1) + \mu (\cup_{n =2} ^{\infty} A_n) = \mu(A_0) + \mu(A_1) + \mu_C (\cup_{n =2} ^{\infty} A_n)$$ because the $\cup_{n =2} ^{\infty} A_n$ results in finitely many disjoint interals, all of them finite. $$\mu(A_0) + \mu(A_1) + \mu_C (\cup_{n =2} ^{\infty} A_n) = \mu(A_0) + \mu(A_1) + \sum _{n =2} ^{\infty} \mu_C (A_n)$$ by countable additivity of $\mu_C$
$$\mu(A_0) + \mu(A_1) + \sum _{n =2} ^{\infty} \mu_C (A_n) = \mu(A_0) + \mu(A_1) + \sum _{n =2} ^{\infty} \mu (A_n)$$ $$= \sum \mu(A_n)$$

So the proof is done because $\mu(\cup A_n) = \sum \mu(A_n)$

share|cite|improve this answer

Your Answer


