Folland contains the following exercise:
Let $F(x) = x^2\sin(\frac{1}{x})$ and $G(x) = x^2\sin(\frac{1}{x^2})$ for $x \neq 0$ and $F(0) = 0 = G(0)$. Show that $F$ and $G$ are differentiable everywhere but $F \in BV([-1,1])$ and $G \notin BV([-1,1])$
(BV is the set of functions of bounded variation)
I'm confused about the point Folland is trying to make with this example. Is he trying to say $F$ and $G$ are differentiable everywhere but $G$ cannot be attained by integrating a function? Doesn't that contradict the fact that I just took derivatives of $G$? I think I'm confusing an important detail, but can't figure out what.
There's no contradiction.
Both $F$ and $G$ are everywhere differentiable, hence each can be obtained by integrating their respective derivatives, but that doesn't imply that they are of bounded variation.
Presumably, that's the point of the exercise.
On a given interval, say $[-1,1]$, an everywhere differentiable function might be of bounded variation (e.g., $F$), or it might not be of bounded variation (e.g., $G$).