# The Countable Additivity of the Lebesgue Premeasure

What is a proof of the countable additivity of the Lebesgue premeasure without using the more general Lebesgue measure?

By the Lebesgue premeasure, I mean the Lebesgue measure restricted to finite unions of almost disjoint rectangles (which may have sides of infinite length).

This condition is used to apply the Carathéodory extension theorem to the Lebesgue premeasure to get the Lebesgue measure.

• What do you mean by "the Lebesgue premeasure"? – Eric Wofsey Aug 6 '18 at 21:05
• @Eric I edited the question to define Lebesgue premeasure. – user109871 Aug 6 '18 at 21:08

The trick is to use compactness for replacing countable additivity by finite additivity.

Let $\mathcal{M}$ be the set of all sets that are unions of finitely many non-overlapping rectangles. We need that:

$~~~$ if $A,B_1,B_2,\ldots\in \mathcal{M}$,

$~~~$ $B_1,B_2,\ldots$ are pairwise disjoint

$~~~$ and $A=\bigcup B_i$,

$~~~$ then $\lambda(A)=\sum\lambda(B_i)$.

I. Proof that $\lambda(A)\ge \sum\lambda(B_i)$: By the finite additivity and monotonicity, for every $N$ we have $$\lambda(A) \ge \lambda\left(\bigcup_{i=1}^N B_i\right) = \sum_{i=1}^N \lambda(B_i).$$ Now take $N\to\infty$.

II. Proof that $\lambda(A)\le \sum\lambda(B_i)$:

The statement is trivial is one of $\lambda(B_i)$ is infinite; so we assume that all $B_i$ are bounded.

Take an arbitrary $c<\lambda(A)$ and $\varepsilon>0$. Take a compact set $K\in\mathcal{M}$ such that $K\subset A$ and $\lambda(K)>c$. For each index $i$, take an open set $G_i\in\mathcal{M}$ such that $G_i\supset B_i$ and $\lambda(G_i)<\lambda(B_i)+\frac{\varepsilon}{2^i}$. Then $$K \subset A \subset \bigcup_{i=1}^\infty B_i \subset \bigcup_{i=1}^\infty G_i,$$ by the compactness of $K$, there is some $N$ such that $$K \subset \bigcup_{i=1}^N G_i.$$ Then $$c < \lambda(K) \le \sum_{i=1}^N \lambda(G_i) \le \sum_{i=1}^N \bigg(\lambda(B_i)+\frac\varepsilon{2^i}\bigg) \le \sum_{i=1}^\infty \lambda(B_i) + \varepsilon$$ so for all $c<\lambda(A)$ and $\varepsilon>0$, $$c < \sum_{i=1}^\infty \lambda(B_i) + \varepsilon.$$ Taking $c\to\lambda(A)$ and $\varepsilon\to+0$ completes the proof.