## IMT Exercise 1.1.2

Give an alternate proof of Lemma 1.1.2(ii) by showing that any two partitions of ${E}$ into boxes admit a mutual refinement into boxes that arise from taking Cartesian products from elements from finite collections of disjoint intervals.

Lemma 1.1.2(ii) If ${E}$ is partitioned as the finite union ${B_1\cup\ldots\cup B_k}$ of disjoint boxes, then the quantity ${m(E)=|B_1|+\ldots +|B_k|}$ is independent of the partition. In other words, given any other partition ${B'_1\cup\ldots\cup B'_{k'}}$ of ${E}$, one has ${|B_1|+\ldots +|B_k|=|B'_1|+\ldots +|B'_{k'}|}$.

The diagram below shows an example with two overlapping boxes, the second column show two different partitions and these can be refined further giving the result shown in the last column.

One way of refining these partitions is suggested below, the lighter boxes are compared and their intersection saved. The remainder(s) are retained for comparison at a later stage although in the illustration there is only on and it is used straight away.

It is clear that the elementary measure is the same for both of these partitions.

Let two partitions of ${E}$ as the finite union of disjoint boxes be ${B_1\cup\ldots\cup B_k}$ and ${B'_1\cup\ldots\cup B'_{k'}}$. We aim to refine these partitions so that at the end the partitions are the same. This will give us our result.

Consider two intersecting boxes, ${B_i}$ and ${B'_{j}}$, if ${B_i=B'_j}$ then no refinement is necessary. Now we focus on the case when ${B_i\supset B'_{j}}$. For ease of notation set ${X=B_i}$ and ${Y=B'_{j}}$ and consider the intervals ${I_{X,1}\ldots I_{X,d}}$ corresponding to box ${X}$ and how they are partitioned by the intersection with ${Y}$. Consider ${I_{X,k}}$ the intersection with ${I_{Y,k}}$ partitions this interval into

• ${I_{X,k}^M}$ the part of the interval ${I_{X,k}}$ which intersects with the corresponding interval of ${Y}$, ${I_{Y,k}}$.
• ${I_{X,k}^L}$ the lower interval, this is the part of the interval ${I_{X,k}}$ which is less than the interval ${I_{Y,k}}$, or just the lower endpoint of ${I_{X,k}}$.
• ${I_{X,k}^U}$ the upper interval, this is the part of the interval ${I_{X,k}}$ which is greater than the interval ${I_{Y,k}}$, or just the upper endpoint of ${I_{X,k}}$.

The boxes formed from the Cartesian products of these intervals form a refinement of ${B_i}$ (which includes ${B'_j}$).

When ${B'_j\supset B_i}$ we use a similar procedure to find a refinement of ${B'_j}$ which contains ${B_i}$.

We can repeat this process, comparing boxes, until no more refinement is possible. As the refinement of a box does not alter its elementary measure we obtain our result, i.e. that ${m(E)=|B_1|+\ldots +|B_k|=|B'_1|+\ldots +|B'_{k'}|}$.

${\Box}$