Exercise 1.7.8 (Hahn-Kolmogorov extension is unique up to a restriction II)

(i)

${\mu }_0$ is a pre-measure.
First, we must demonstrate that $A$ is a Boolean algebra.

• We have $\varnothing =[0,0)\in A$.

• Let $[a,b),[c,d)\in A$. We have to demonstrate that their set difference is also an interval of the form $[k_1,k_2)$ for some $k_1,k_2\in ℝ$. To prove this, we need to apply a brute-force case-by-case proof. We will not perform this easy albeit tedious exercise here; instead, we leave a graphic sketch of all 6 non-trivial (i.e., $a<b$ and $c<d$) cases.

  The assertion about the set difference of two unions of half-open intervals follows directly by De-Morgan’s laws.

• Closeness under finite unions follows directly from the definition.

Next we demonstrate that ${\mu }_0:A\to [0,+\infty ]$ is a pre-measure.

• We have ${\mu }_0(\varnothing )=0$ by definition.

• For any sequence ${(E_n)}_{n\in \mathbb{N}}$ of disjoint $A$-measurable sets, we have

  $${\mu }_0\left(\bigcup _{n=1}^{\infty }{E}_n\right)\ge {\mu }_0(E_j)=\infty \sum _{n=1}^{\infty }{\mu }_0(E_n)\ge {\mu }_0(E_n)=\infty$$

  whenever there is at least $E_j\ne \varnothing$ (otherwise the claim is trivial as well).

(ii)

$⟨A⟩$ is the Borel $\sigma$-algebra $B[R]$.
Recall from Exercise 1.4.14 (Generators of the Borel $\sigma$-algebra) that to show that two sets, the set $A$ of all half-open intervals of $ℝ$ and the set $O$ of all open intervals of $ℝ$, generate the same $\sigma$-algebra, it suffices to show that every $\sigma$-algebra containing $A$ contains $O$ and vice versa.
Thus, let $S$ be a $\sigma$-algebra containing $A$. Let $[a-𝜖,b)\in O$ for any $𝜖>0$ and thus $$\bigcup _{n=1}^{\infty }\left[a-\frac{1}{n},b\right)=(a,b)\in S$$

as well. Conversely, let $S$ be a $\sigma$-algebra containing $O$ and let $[a,b)\in A$ arbitrary. Then, $(a+𝜖,b)$ is contained in $O$ for every $𝜖>0$ and so

$$\bigcap _{n=1}^{\infty }\left(a+\frac{1}{n},b\right)=[a,b)\in S$$

too.

(iii)

The Hahn-Kolmogorov extension $\mu :B[R]\to [0,+\infty ]$ of ${\mu }_0$ assigns an infinite measure to any non-empty Borel set.
This is obvious, as the Hang-Kolmogorov extension is the restriction of the outer measure which is the infimum over infinities itself.

(iv)

Show that counting measure $\#$ (or more generally, $\mathit{c\#}$ for any $c\in (0,+\infty ]$) is another extension of ${\mu }_0$ on $B[R]$. Any non-empty half-open interval $[a,b)$ contains uncountably many points and thus $\#([a,b))=\infty$ by definition. For empty sets $\#(\varnothing )=0$. Thus, $\#$ and ${\mu }_0$ agree on $A$.

In other words, the finite sets

are contained in $B$ but not in the Borel algebras. Inside the Borel algebra of half-open intervals, the measures $\mu$ and $\#$ coincide, yet on the countable sets $\{a_1,\dots ,a_n\}$ they disagree as the Hahn-Kolmogorov extension assigns them a measure of $\infty$ yet $\#$ gives them the measure of $0$.