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.
 A: 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. 
A: Well,first of all,your definition of integrability isn't quite right-although it's a subtle mistake even experts make sometimes. Taking P as a given partition of the real line in the domain of f, the definition you're looking for is the informal one: $\lim_{|P|\rightarrow 0} U(P)-L(P) = 0$. It should be noted this isn't really a precise definition of integrability since P is a partition and not really a variable in the usual sense. But the basic idea is that the upper and lower Riemann sums converge to the same value if the limit of the mesh of P goes to 0. 
The idea in the proof is to relate the absolute value the function f on the partition P of I =[a,b]to f the same way the absolute value function on R relates positive and negative real numbers. Consider the definition of absolute value: 
$ |x| =
\begin{cases}
\ - x  & x < 0 \\
  x, & x\geq 0
\end{cases}$
We now need an analogous construction on the partition P for f.Define the following 2 functions on R:
(1) $f^{+}(x) =
\begin{cases}
\ f(x)  & f(x)\geq 0 \\
  0, & f(x)< 0
\end{cases}$
$f^{-}(x) =
\begin{cases}
\ - f(x)  &  f(x)\leq 0 \\
  0, & f(x)> 0
\end{cases}$
Note the similarity in the definitions to the absolute value function definition. We'll now use this definition to prove that
(2) f(x) is integrable on [a,b] implies that $f^{+}(x)$ is integrable on [a,b]. 
It's clear by these definitions that $f^{-}(x)$ =-($(-f(x))^{+}$). This allows us to not only determine that $f^{-}(x)$ is integrable, by the fact that a constant multiple of any integrable function f is integrable. Then since the sum of 2 integrable functions is integrable on the same partition:
(3)|f(x)|=  $f^{+}(x)-f^{-}(x)$ 
All that's left to prove is (2).(3) then gives the desired result. To prove (2),if f is bounded on the interval I= [a,b},we need to show the following:
(4) sup( $f^{+}(x)$) - inf($f^{+}(x)$) $\leq$ sup(f(x)) - inf(f(x))  
You should try and prove (4) yourself. Hint: You have to test each case using the definition (1) and the cases $f(x)\leq 0$ and $f(x)\geq 0$. If should be clear from the definition of the lower and upper Riemann sums that if we prove (4), this proves (2).From the fact $f^{-}(x)$ is then integrable and the sum of 2 integrable functions is integrable, this proves |f(x)| is integrable on I and we're done! 
Now do you understand why in calculus your professors frowned when you asked WHY for anything? lol 
A: There is a much more general result here: If $f:[a,b] \to [c,d]$ is Riemann integrable and $g$ is continuous on $[c,d],$ then $g\circ f$ is Riemann integrable on $[a,b].$ (In your problem we would take $g(x) = |x|.$)
