How do we Riemann-integrate over subsets of the real line? Suppose we're high school teachers standing in front of a group of intelligent, logical and intellectually voracious 17 year olds. We've already defined the meaning of the notation $$\int_a^b f(x)dx$$ as a Riemann integral, e.g. using Darboux sums, and the students have learned how to compute some basic integrals using FTC.
We now wish to move to probability theory. If $f$ is the density function of some continuous probability distribution $\mathbf{P}$ on the real line, then for all sufficiently nice $A \subseteq \mathbb{R}$, we have $$\mathbf{P}(A) = \int_{x \in A} f(x)dx$$
Notice that this is an integral over a subset of the real line. So, we want to get down a definition that allows us to Riemann-integrate over subsets of the real line. The kids are smart, so the definition can be pretty sophisticated. But Lebesgue integrals are currently (just) out of reach, so it has be a Riemann-style definition.

Question. How do we Riemann-integrate over subsets of the real line?

 A: First, extend the definition of the integral to cover integration over $( -\infty, \infty )$ by taking limits over compact intervals:
\begin{equation}
\int\limits_{-\infty}^{\infty}f(x) \text{d}x = \lim \limits_{n \to \infty} \int\limits_{a_n}^{b_n}f(x) \text{d}x
\end{equation}
where $[a_n, b_n]$ is some exhaustion by compact sets (colourful language is always fun).
Then prove that the above is independent of the particular exhaustion for suitably nice functions.  
EDIT: This point is crucial, because unlike compact intervals, continuity of $f$ is not enough to obtain a unique limit (think about principal values). You have to make clear that $f$ needs to decay fast enough at infinity (which is easy enough to picture though).
Now define
\begin{equation}
\int\limits_{x \in A}f(x) \text{d}x = \int\limits_{-\infty}^{\infty} \chi_A(x) \cdot f(x) \text{d}x
\end{equation}
for subsets $A \subseteq \mathbb{R}$, where $\chi_A$ is defined as
\begin{equation}
\chi_A(x) = \left\{
     \begin{array}{lr}
       1 & : x \in A\\
       0 & : x \notin A
     \end{array}
   \right.
\end{equation}
Note, that you now have two definitions for intergration over intervals. So prove that they are equivalent.  
Also give nice counterexamples where any of the above procedures fail, emphasizing the need of $f$ to be $\textit{suitably nice}$.
