Length is a Pre-Measure on the Semi-Ring of Bounded Half-Open Real Intervals

Question

I am looking for a simple proof that length $l$ is a positive pre-measure on the semi-ring of bounded half-open intervals and would appreciate feedback on my following effort........

By definition $l(\emptyset) = 0 $, and for a "proper" bounded half-open interval $I = [a, b): b > a$ then $l(I) = b - a > 0$ and $l$ is a positive function. So that leaves countable additivity.

Let $\{I_j\}_{j = 1, \infty}$ be a countable set/sequence of disjoint half open intervals: $I_j = [a_j, b_j)$ whose union is a bounded half-open interval $I = [a, b)$. Then....

1. We can assume that $\{a_j\}_{j = 1, \infty}$ are ordered $a = a_1 < a_2 < .....$. This follows since $\{a_j\}_{j = 1, \infty}$ is bounded below by $a$, must be distinct since the intervals are disjoint, and must contain a least element since $[a, b)$ is left closed.
2. For all $j$, $a_{j+1} = b_j$. If not and $b_j < a_{j+1}$ then $[b_j , a_{j+1}) \not \subset I$ and therefore $I$ is not an interval, while if $a_{j+1}< b_j $ then $ I_j \cap I_{j+1} = [a_{j+1}, b_j) \ne \emptyset$ which contradicts $\{I_j\}_{j = 1, \infty}$ being disjoint.
3. For the first $n$ intervals, using (2), $\cup_{j = 1, n} I_j$ is an interval $[a_1, b_n)$ and $l(\cup_{j = 1, n} I_j) = b_n - a_1 = (b_1 - a_1) + ( b_2 - b_1+ + .... (b_n - b_{n-1}) = \Sigma_{j = 1, n} l(I_j)$.
4. By definition, $\Sigma_{j = 1, \infty} l(I_j)$ = Lt$_{n \to \infty} \Sigma_{j = 1, n} l(I_j)$ if the limit exists. And Lt$_{n \to \infty} \Sigma_{j = 1, n} l(I_j) = $ Lt$_{n \to \infty} b_n - a_1$. Since $\{b_n\}$ is monotonic and bounded with $b = sup(b_n)$ the limit does exist and $\Sigma_{j = 1, \infty} l(I_j)$ $= b - a_1 = b - a = l(\cup_{j = 1, \infty} I_j)$. So, $l $ is countably additive.

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After further thought I realise that this is not a correct proof. I'll leave it here in case it helps anyone. It fails at step (1). The seemingly obvious claim that a (distinct) countable set of reals with a minimum element can be put in ascending sequence is wrong. After failing to formally prove this I looked for a counterexample and quickly found $A = \{0\} \cup \{1/n: n \in \mathbb N^+ \}$. In ascending sequence clearly $a_1 = 0$ but for any choice of $a_2 = 1 /n$ there is $0 < 1/(n+1) < a_2$.

So, I'm back to reading Halmos - Measure Theory pp.32-35.

A further counterexample with intervals is $\mathscr I = \{[1-1/n, 1-1/(n+1)):n \in \mathbb N^+ \} \cup \{[2-1/n, 2-1/(n+1)):n \in \mathbb N^+ \}$. The intervals are disjoint with union = $[0, 2)$, Clearly in an ascending sequence $a_1 = [0, 1/2)$ and $a_{100} = [99/100, 100/101)$ , but where for example is $[1, 3/2)$ ?

Answer 1

The most simple proof I can come up with uses (sadly) the Heine-Borel theorem.

For $n \in \mathbb{N}$, the one inequality $\sum_{j=1}^n b_j - a_j \leq b-a$ indeed follows by ordering the intervals $a_{j_1}\leq b_{j_1} \leq \cdots \leq a_{j_n}\leq b_{j_n}$ and making a telescoping sum.

For the other inequality, take $\epsilon > 0$, then $N$ many intervals $(a_j - \epsilon/2^j, b_j)$ cover the interval $[a,b-\epsilon]$ (let's say the first $N$, which is the worst case scenario).

Then the desired inequality follows by writing $$b- \epsilon - a \leq \sum_{j=1}^N b_j - a_j + \epsilon/2^j \leq \sum_{j=1}^\infty b_j - a_j + \epsilon.$$