How to prove that $C = \bigcup_{a \in A \times \{0\}} \{(1-t)a+tv| 0\leq t \leq 1\}$ is Jordan-measurable if $A$ is Jordan-measurable? Let $A \subseteq \mathbb{R}^{n-1}$ be Jordan-measurable and
$$C = \bigcup_{a \in A \times \{0\}} \{(1-t)a+tv| \; 0\leq t \leq 1\}$$
be an object in $\mathbb{R}^n$ with the base area $A \times \{0\}$ and the apex $v \in \mathbb{R}^n$. I have to show that $C$ is Jordan-measurable and $\mu_n(C) = \frac{|v_n|}{n} \mu_{n-1}(A)$. Now our assistant professor told us that one usually proves "such" propositions in three steps:


*

*For every cuboid $A$.

*For $A$ being a finite union of disjoint cuboids $Q_i$.

*For every Jordan-measurable $A$.


Now I don't understand how this helps or more specific: How do I apply this strategy? What exactly do I want to prove for every cuboid $A$ (and for $A$ being a finite union of disjoint $Q_i$ and eventually for every Jordan-measurable $A$)? Also, why does this strategy work? What idea is there behind this strategy?
Thanks for any help in advance.
 A: "What exactly do I want to prove ... eventually for every Jordan-measurable $A$?"  I
"I have to show that $C$ is Jordan-measurable and $\mu _n(C) = \frac{|v_n|}{n}\mu _{n-1}(A)$."  
A good way to start is always to review the definitions and make sure you've read the problem carefully.  You may have problems solving the problem and getting things to click simply because the definitions aren't fresh in your mind at the time, or you're not perfectly clear on what the problem's asking. 
Okay, so let's use $C_A$ to denote your $C$ above, since its definition depends on $A$.  Now if $Q$ is a cuboid, how would you prove this?  It may be easier to think about in a special case.
HINT 1:

 Often a good idea is to solve the problem in the lowest possible dimension in which it makes sense.  If $Q$ were a cuboid in $\mathbb{R}$ with length $b$, i.e. an interval, and $v = (x,h)$ is a vector in $\mathbb{R}^2$, then $C_Q$ is a triangle with base $b$ and height $|h|$.  How do you show its Jordan measure is $\frac{1}{2}bh$?  Now generalize.

If you've proven it for cuboids $Q$, it should be almost immediately obvious for $D = Q_0 \sqcup \dots \sqcup Q_k$ a disjoint union of cuboids.
HINT 2:

 Since $C_D$ = $C_{Q_0} \sqcup \dots \sqcup C_{Q_k}$

Finally, how would you prove it for arbitrary $A$?  What's the connection between Jordan measure of a set and unions of cuboids?
HINT 3:

 Lemma: The outer measure of a set $S$ is defined to be the infimum of the measures of the simple sets (finite unions of cuboids) which contain $S$, but you can prove that the outer measure is the infimum of the measures of the measurable sets which contain $S$.  Obviously this can't replace the definition since measurability is defined in terms of outer measure, so it would give a circular definition.  Still, this lemma can be useful.

HINT 4:

 Lemma: The outer measure of a set $S$ is defined to be the infimum of the measures of the simple sets which contain $S$, but you can prove that the outer measure is the infimum of all the finite-disjoint-unions-of-cuboids that contain it.

