Lebesgue Pre-Measure is Pre-Measure

Theorem
The Lebesgue pre-measure $\lambda^n$ on the half-open $n$-rectangles $\JJ_{ho}^n$ is a pre-measure.

Proof
We employ Characterization of Pre-Measures.

It is known that $\map {\lambda^n} \O = 0$ by definition of Lebesgue pre-measure.

The only possibility for two disjoint half-open $n$-rectangles to constitute a single, large half-open $n$-rectangle is when they are of the form:


 * $\horectr {\mathbf a} {\mathbf b} \quad \horectr {\mathbf a'} {\mathbf b'}$

such that we have for some $i$ with $1 \le i \le n$:


 * $i \ne j \implies a_j = a'_j$
 * $i \ne j \implies b_j = b'_j$
 * $i = j \implies a'_j = b_j$

which intuitively can be visualised as two cubes that together form one large bar, namely $\horectr {\mathbf a} {\mathbf b'}$.

In this situation, we have:

Thus it is verified that $\lambda^n$ is finitely additive.

Finally, suppose that $\horectr {\mathbf a_m} {\mathbf b_m} \downarrow \O$ is a decreasing sequence of sets, with limit $\O$.

Then there exists at least one $j$ with $1 \le j \le n$ such that:


 * $\ds \lim_{m \mathop \to \infty} a_{m, j} = \lim_{m \mathop \to \infty} b_{m, j}$

which by Combination Theorem for Sequences is equivalent to:


 * $\ds \lim_{m \mathop \to \infty} b_{m, j} - a_{m, j} = 0$

The fact that the sequence is decreasing means that, from Cartesian Product of Subsets, for all $m \in \N$, for all $1 \le i \le n$:


 * $\hointr {a_{m, i} } {b_{m, i} } \subseteq \hointr {a_{1, i} } {b_{1, i} }$

and whence $b_{m, i} - a_{m, i} \le b_{1, i} - a_{m, 1}$.

Hence we have:

This verifies the last condition for Characterization of Pre-Measures, since $\lambda^n$ only takes finite values.

Hence $\lambda^n$ is a pre-measure.