What is the motivation/applications for the definition of Lebesgue measure on $\mathbb R^n$? The definition of the Lebesgue measure on $\mathbb R^n$ is fundamentally tied up with the following assumption:

The measure of the cartesian product of $n$ intervals should be the product of the lengths of those intervals.

That is to say, to define measure on $\mathbb R^n$, we just blindly generalize the formula for the area of a rectangle or the volume of a cuboid. Why do this?
Cartesian products of intervals, and the concept of multiplying their lengths together, are only relevant in $\mathbb R^2$ and $\mathbb R^3$. In fact, even there, they're only relevant when those sets are being interpreted as models for physical space. If I was working on a two-variable problem in which the variables were temperature and pressure, then measuring the "size" of the set of all configurations with temperature in $[0, 2]$ and pressure in $[5, 3]$ (say) by multiplying together $|2-0|$ and $|5-3|$ would seem like a very irrelevant thing to do. And if I'm working on an $n$-variable problem, then claiming that this operation could possibly provide a useful notion of the "size" of a set just seems absurd.
 A: You want to develop the Lebesgue integral in a way which generalises the Riemann integral. In order to do that, you need to specify the Lebesgue measure. Regardless of how you want to define Lebesgue measure $m$, for any set $A \subseteq \mathbb{R}^n$ that you want to measure, it should satisfy 
$$m(A) = \int_{\mathbb{R}^n}\chi_A dm$$
where the integral on the right is the Lebesgue integral of the characteristic function of $A$. But hold on, if we're pretending like we don't have the Lebesgue measure at our disposal, what does the Lebesgue integral even mean? Well, all I've said about it so far is that is should generalise the Riemann integral, at least for nice functions.
Now consider $A = [a_1, b_1]\times\dots\times[a_n, b_n] \subseteq \mathbb{R}^n$. This is a pretty reasonable set (it's a Borel set after all), so you would hope that the Lebesgue measure of this set is defined. Furthermore, $\chi_A$ is a reasonable function (it is Riemann integrable), so you'd hope that you can make sense of its Lebesgue integral. But if we want the Lebesgue integral to generalise the Riemann integral, they should give the same number. That is,
$$m(A) = \int_{\mathbb{R}^n}\chi_A dm = \int_{\mathbb{R}^n}\chi_A dx.$$
But we know how to evaluate the last integral:
$$\int_{\mathbb{R}^n}\chi_A dx = \int_{a_n}^{b_n}\dots\int_{a_1}^{b_1}1dx_1\dots dx_n = (b_1 - a_1)\dots(b_n - a_n).$$
So defining $m([a_1, b_1]\times\dots\times[a_n, b_n]) = (b_1 - a_1)\dots(b_n - a_n)$ seems pretty reasonable if you want the Lebesgue integral to generalise the Riemann integral.
