# Verify conditions of the Extension Theorem for a probability measure $\mathbb P$ of the Uniform[0,1] distribution

I am (self-)studying the book by Rosenthal called A first look at rigorous probability theory. My question is on verifying the conditions on a probability measure $\mathbb P$ of the Uniform[0,1] distribution such as to apply the Extension Theorem (Theorem 2.3.1, page 10). More specifically, I would like to solve Exercise 2.4.3 on page 15. Definitions and variables used are all explained on page 15.

Let me highlight that interval'' is understood to include all the open, closed, half-open, and singleton intervals contained in $[0,1]$, and also the empty set $\emptyset$.

(a) Let $a_j$ be the left end-point and $b_j$ the right end-point of the interval $I_j$ and similary $a_0$ and $b_0$ for the interval $I$. By re-ordering we can ensure $a_0 \ge a_1 \le b_1 = a_2 \le \ldots \le b_k \ge b_0$. Thus, \begin{align} \sum_{j = 1}^n \mathbb P(I_j) = \sum_{j = 1}^n (b_j - a_j) = b_k - a_1 \ge b_0 - a_0 = \mathbb P (I). \end{align}

(b) $I_1,I_2,\ldots$ is a countable collection of open intervals with $\bigcup_{j=1}^\infty I_j \supseteq I$ for some interval $I$. Using the Heine-Borel Theorem we have that $\exists k : \bigcup_{j = 1}^k I_j \supseteq I$. How to continue from here?

(c) I do not know how to solve this part.

• Rosenthal. 
– Did
Apr 24, 2014 at 9:40
• My bad, small typo. Thanks!
– Ritz
Apr 24, 2014 at 9:47

For (b), one we got the existence of $k$ such that $I\subseteq \bigcup_{j=1}^k I_j$, we use (a) to obtain $\mathbb P(I)\leqslant\sum_{j=1}^k\mathbb P(I_j)\leqslant \sum_{j=1}^{+\infty}\mathbb P(I_j)$.

(c) We have to show that if $I$ is an interval and $(I_j)_{j\geqslant 1}$ is a collection of intervals such that $I\subset\bigcup_{j=1}^\infty I_j$, then $\mathbb P(I)\leqslant\sum_{j=1}^{+\infty}\mathbb P(I_j)$. Here no assumption of openness or closeness is done. We try to use the previous case and we enlarge each $I_j$ "but not too much": for a fixed $\varepsilon$, define $I_j^\varepsilon:=(a_j-\varepsilon 2^{-j},b_j+\varepsilon 2^{-j})$ if $I_j=\cdot a_j,b_j\cdot$, where the left $\cdot$ may denote $($ or $[$ and similarly for the right. If $I=(a,b)$, then define $I^\varepsilon:=[a+\varepsilon,b-\varepsilon]$. Then for each positive $\varepsilon$, $$I^\varepsilon\subset \bigcup_{j\geqslant 1}I_j^\varepsilon, \quad I^\varepsilon\mbox{ is closed }, \quad I_j^{\varepsilon}\mbox{ is open}.$$ By (b), we deduce that $\mathbb P(I^\varepsilon)\leqslant \sum_{j\geqslant 1}\mathbb P(I_j^{\varepsilon})$. Since $\mathbb P(I^\varepsilon)=\mathbb P(I)-2\varepsilon$ and $\sum_{j\geqslant 1}\mathbb P(I_j^{\varepsilon}) =\sum_{j\geqslant 1}\mathbb P(I_j)+\varepsilon$, we obtained for each positive $\varepsilon$, $$\mathbb P(I)\leqslant \sum_{j\geqslant 1}\mathbb P(I_j)+3\varepsilon.$$

• Did I answer (a) correctly as well? I think I used that the intervals are disjoint (which they are not necessarily).
– Ritz
Apr 24, 2014 at 9:49

I don't have enough reputation to comment, so I am attempting to answer:

I don't think (a) is correct because, as you pointed out, we cannot assume the intervals are disjoint. Here is my attempt at a correct solution to (a). Note that I may very well be incorrect, but what I try to do is break the intervals into disjoint intervals and add back in the overlap (which should work because the uniform probability list assigns probability based on on the "length" of intervals). Here is a rough outline/ proof sketch.

Let $I_{1} , I_{2} \dots I_k$ be a finite collection of intervals with $\cup_{j=1}^{k} I_j \supseteq I$ for some interval I.

For $1 \leq j \leq k$, write $a_j$ for the left end point of $I_j$, $b_j$ for the right endpoint of $I_j$. We can re-order the interval so that $a_1 \leq a_2 \leq \dots \leq a_k$. If $b_i \geq b_{i+1}$ define $c_i$ s.t. $b_i - c_i = a_{i+1}$.

Then we have $a_1 \leq b_1 - c_1 = a_2 \dots \leq b_k - c_k$ with $c_k = 0$ (since there is no $a_{k+1}$). Letting $I$ be the interval from $a_0$ to $b_0$, $a_1 \leq a_0 , b_0 \leq b_k - c_k$. Thus, $\sum P(I_j) = \sum_j (b_j - a_j) = \sum_j ((b_j - c_j) - a_j) + c_j = \sum_j ((b_j - c_j) - a_j) + \sum_j c_j = (b_k - c_k) - a_1 + \sum_j c_j$.

Since $b_k \geq b_0 , a_1 \leq a_0, \sum_j c_j \geq 0$, $(b_k - c_k) - a_1 + \sum_j c_j \geq P(I)$