Conditions for Anti-Differentiability Taking the anti-derivative (or integral) of a function can be a difficult problem. What are some conditions that ensure that a function is anti-differentiable, or that show that it is not anti-differentiable? 

It will take you a few years to get to this. A bounded function is Riemann integrable, on a closed interval, if and only if the set of points at which it is discontinuous has measure zero. This result is probably due to Lebesgue, let me look up some things. In Royden, exercise 2b on page 82. In Berberian, page 268, Theorem 5.13.7. 
https://en.wikipedia.org/wiki/Riemann_integral#Integrability
The standard sort of example is this: taking, say, $0 \leq x \leq 1,$ whenever $x$ is irrational define $f(x) = x.$ However, when $x = p/q$ is a fraction in lowest terms, meaning integers $p,q > 0$ and $\gcd(p,q) = 1,$ then take
$$ f \left( \frac{p}{q} \right) = p \sin \left( \frac{1}{q} \right).  $$
This function $f$ is discontinuous at every rational number. However, fiddling a bit shows that $f$ is continuous at all irrational numbers. That is, in order for a rational number to get very close to a fixed irrational number $\alpha,$ it is necessary for the denominator $q$ to become very large, which then implies that $f(p/q)$ is quite close to $p/q$ and therefore quite close to $\alpha.$ Modest use of Diophantine approximation. So, the thing is Riemann integrable. So There. 
