Conditions for a function f: R-->R (Reals) to have an antiderivative. Ok, so I gather if a function $f: \mathbb R \rightarrow \mathbb R $  is to have an antiderivative. 
The first/obvious guess is to use F(x) with :
$$F(x):= \int_a^x f(t)dt +C; C $$ a Real constant.
Then the FTC grants us : 
     $$ F'(x)=f(x) $$
Now, the details:
  1) I believe f(x) must be continuous at $x$ in order to have $F'(x)=f(x)$, right?
2)Must the (Riemann) integral converge for the identity to apply?
  If so, then f must be a.e. continuous and bounded.
3)Curious: Since F is differentiable, it is continuous. So we then
   need F to be continuous on x (t is just a dummy variable, so it won't matter). For this last it seems boundedness of $f$ is enough, in that
  $F(x+h)-F(x):= \int_x^{x+h} f(t)dt$ can be made small-enough by making
  $h$ small-enough (right?) giving us a $\delta - \epsilon$ argument for the
  continuity of $F$ at $x$
  Can this be extended to $\mathbb R^n$? I am thinking of conservative
  vector fields and the condition $U_x= V_y$ for $n=2$ but I can't see
  for $n>2$. 
   Is this  right?
    TIA.
  EDIT: I am trying to determine the most general conditions under which
 $F'=f$
   Thanks.  
 A: *

*Wrong. If $f$ is continuous, then $F'=f$. But in some cases, you still have $F'=f$ even if $f$ is discontinuous.

*There are no convergence questions here. In order that this makes sense, $f$ must be Riemann-integrable. And then $\int_a^xf(t)\,\mathrm dt$ exists automatically.

A: For the Riemann integral of $f$ to exist on a interval, you need that the set of discontinuity points of $f$ has zero Lebesgue measure. In particular, if the set of discontinuity points of $f$ is countable (or finite, of course), then the Riemann integral of $f$ exists. 
In the above situation, i.e. when $f$ is Riemann-integrable on an interval $[a,b]$, then $F(x)=\int_a^x f(t)\,dt$ exists and $F'(x)=f(x)$ at all points of continuity of $f$. Where $f$ is not continuous, you can have things like this: let $f=1_{[1,2]}$, that is $f(x)=1$ on $[1,2]$ and $0$ elsewhere. Then, as $f$ has a jump at $1$, it cannot be the derivative of a function on intervals $(1-\delta,1+\delta)$ (because it fails Darboux's Theorem). 
