If f is integrable, why is |f| integrable? For a function to be integrable, wouldn't it by definition have to mean that the upper sum $ U(f) $ and lower sum $ L(f) $ exist such that
$ U(f) - L(f) = 0 $? 
So regardless of what $ U(f) - L(f) $ we know that 
$ U(f) - L(f) \leq |U(f) - L(f)|= 0 $.
So by default, wouldn't $ |f| $ just be integrable like $ f$? I suppose I understand why, but I'm just not sure how to prove it.

If $U(f,P) - L(f,P) < \epsilon$, then we also have $U(|f|,P) - L(|f|,P) < \epsilon$ since, in particular, $\max(|M_i|, |m_i|) - \min(|M_i|, |m_i|) \leq M_i - m_i$ in every subinterval. 
Intuitively, this means that taking the absolute value of $f$ has the effect of reducing (or simply not changing) the magnitude of the oscillation on a given subinterval. Integrability is thus preserved since the upper and lower sums can only get "closer" to each other if one takes the absolute value, never farther. 
