Lebesgue Pre-Measure is Pre-Measure

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Theorem

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


Proof

We employ Characterization of Pre-Measures.


It is known that $\lambda^n \left({\varnothing}\right) = 0$ 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:

$\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \quad \left[[{\mathbf a' \,.\,.\, \mathbf b'}\right))$

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 $\left[[{\mathbf a \,.\,.\, \mathbf b'}\right))$.


In this situation, we have:

\(\displaystyle \lambda^n \left({\left[\left[{\mathbf a \,.\,.\, \mathbf b}\right)\right)}\right) + \lambda^n \left({\left[\left[{\mathbf a' \,.\,.\, \mathbf b'}\right)\right)}\right)\) \(=\) \(\displaystyle \prod_{j \mathop = 1}^n \left({b_j - a_j}\right) + \prod_{j \mathop = 1}^n \left({b'_j - a'_j}\right)\) By definition of Lebesgue pre-measure
\(\displaystyle \) \(=\) \(\displaystyle \left({b_i - a_i + b'_i - a'_i}\right) \prod_{\substack{j \mathop = 1 \\ j \mathop \ne i} } \left({b_j - a_j}\right)\) By the noted properties of $a_j, b_j, a'_j, b'_j$
\(\displaystyle \) \(=\) \(\displaystyle \lambda^n \left({\left[\left[{\mathbf a \,.\,.\, \mathbf b'}\right)\right)}\right)\) By definition of Lebesgue pre-measure

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


Finally, suppose that $\left[[{\mathbf a_m \,.\,.\, \mathbf b_m}\right)) \downarrow \varnothing$ is a decreasing sequence of sets, with limit $\varnothing$.

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

$\displaystyle \lim_{m \to \infty} a_{m,j} = \lim_{m \to \infty} b_{m,j}$

which by Combination Theorem for Sequences is equivalent to:

$\displaystyle \lim_{m \to \infty} b_{m,j} - a_{m,j} = 0$

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

$\left[{a_{m,i} \,.,\,.\, b_{m,i}}\right) \subseteq \left[{a_{1,i} \,.\,.\, b_{1,i} }\right)$

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


Hence we have:

\(\displaystyle \lim_{m \to \infty} \lambda^n \left({\left[\left[{\mathbf a_m \,.\,.\, \mathbf b_m}\right)\right)}\right)\) \(=\) \(\displaystyle \lim_{m \to \infty} \prod_{i \mathop = 1}^n \left({b_{m,i} - a_{m,i} }\right)\) By definition of Lebesgue pre-measure
\(\displaystyle \) \(\le\) \(\displaystyle \lim_{m \to \infty} \left({b_{m,j} - a_{m,j} }\right) \prod_{\substack{i \mathop = 1 \\ i \mathop \ne j} } \left({b_{1,i} - a_{1,i} }\right)\) By the above discussion
\(\displaystyle \) \(=\) \(\displaystyle 0\) Combination Theorem for Sequences


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

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

$\blacksquare$


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