A necessary condition to $F'(x)=f(x)$ for a continuous function $f$ 
Theorem:  Consider ,
$$F(x)=\int_a^xf(t)\,dt$$
If the function $f:[a,b]\to \mathbb R$ is continuous then , $F(x)$ is differentiable and $F'(x)=f(x).$

I know that the continuity condition of $f$ is sufficient condition.
That means there exists a discontinuous function $f$ for which this $F'(x)=f(x)$.
My Question:
Does there exist a necessary condition for this ?
$$OR$$
After imposing which extra condition on $f$  it is necessary that $F'(x)=f(x)$ ?
 A: $\newcommand{\Reals}{\mathbf{R}}$To speak of "the converse" of the theorem (as currently written) is ambiguous, because


*

*It's not specified whether $x$ refers to a single number ("if $f$ is continuous at $x$, then $F$ is differentiable at $x$...") or a general number ("for all $x$ in $(a, b)$...");

*No hypotheses for $f$ have been given before the definition of $F$.
The first omission is relatively minor; we can alway focus attention at one point. But the second (failure to specify "standing hypotheses" about $f$) is disastrous in terms of reading meaning into the notion of a "converse". (The linguistic morals include: "Only speak of a converse when you have a well-formed logical implication", and "avoid pronouns such as this and it, particularly when they refer to a logical proposition''. Pronouns already induce plenty of confusion when they refer to some mathematical object in situations with several objects under discussion.)

Based on context gleaned from the precise wording of the question and from subsequent comments, I assume the following wording: Let $a < b$ be real numbers, $f:[a, b] \to \Reals$ a Riemann integrable function, and $F:[a, b] \to \Reals$ the function defined by
$$
F(x) = \int_{a}^{x} f(t)\, dt.
$$
Theorem: If $f$ is continuous on $[a, b]$, then $F$ is differentiable on $(a, b)$, and $F'(x) = f(x)$ for all $x$ in $[a, b]$ (with "differentiability" being one-sided at the endpoints).
Consequently, I assume "the converse" reads: If $F$ is differentiable on $[a, b]$ and if $F'(x) = f(x)$ for all $x$ in $[a, b]$, then $f$ is continuous.
The converse is false. To see why, pick your favorite differentiable (but not continuously-differentiable) function, namely
$$
F(x) = \begin{cases}
  x^{2} \sin(1/x) & x \neq 0, \\
  0 & x = 0.
\end{cases}
$$
The (discontinuous) derivative
$$
F'(x) = \begin{cases}
  2x\sin(1/x) - \cos(1/x) & x \neq 0, \\
  0 & x = 0,
\end{cases}
$$
is bounded and continuous everywhere except $0$, so $f = F'$ is Riemann integrable on an arbitrary closed (bounded) interval.
To be completely explicit, let $[a, b] = [0, 1]$, and put
$$
f(t) = \begin{cases}
  2t\sin(1/t) - \cos(1/t) & 0 < t \leq 1, \\
  0 & t = 0,
\end{cases}\qquad
F(x) = \int_{0}^{x} f(t)\, dt =  \begin{cases}
  x^{2} \sin(1/x) & x \neq 0, \\
  0 & x = 0.
\end{cases}
$$
The function $F$ is differentiable on $[0, 1]$ and $F'(x) = f(x)$ for all $x$ in $[0, 1]$, but $f$ is not continuous. (It should be equally clear we can arrange worse behavior, e.g. $f$ having a dense set of discontinuities.)
The preceding is an attempt to address the question in the comment. As to the questions in the body of the post:


*

*"Is there a necessary condition for this?" Presumably the intended question reads "Is there a (non-vacuous) condition on $F$ or $F'$ guaranteeing $f = F'$ is continuous?" Modulo interpretation, "no". Continuity of $f$ is equivalent to...continuity of $F' = f$.

*"After imposing extra condition[s] on $f$ is it necessary (i.e., logically implied) that $F' = f$?" As demonstrated by the example, that's tricky. In detail, let $c$ be a non-zero real number, and define
$$
g(t) = \begin{cases}
  2t\sin(1/t) - \cos(1/t) & 0 < t \leq 1, \\
  c & t = 0.
\end{cases}
$$
If $G$ denotes the definite integral of $g$ from $0$, then $F = G$, but of course $G'(0) = 0 \neq c = g(0)$.
Philosophically, the definite integral "knows the correct value of $G'(0)$" even though $G'$ is discontinuous. Any extra condition on $f$ has to detect this type of behavior.
To put it more bluntly, suppose you came across the question: Suppose $g(t) = 2t\sin(1/t) - \cos(1/t)$ for $t > 0$. What is the value of $g(0)$? (I can almost guarantee that if this were posted on Math.SE, there'd be a flame-fest, because there are infinitely many answers under the stated conditions.) According to the definite integral / antiderivative, however, there is a unique "correct" answer, $g(0) = 0$. (!) The point is, any extra condition imposed on $g$ has to be able to give this answer. Offhand, I don't see any non-trivial condition, i.e., any condition other than "Integrate, differentiate the integral, and see if the values match".
A: Let $f$ be Riemann integrable on $[a, b]$. Then the function $$F(x) = \int_{a}^{x}f(t)\,dt$$ is known to be continuous and of bounded variation on $[a, b]$. Apart from these properties of $F$ it is possible to ensure differentiability of $F$ at points where $f$ is continuous.
Another point to note is that since $F(x)$ is of bounded variation it possesses a derivative almost everywhere i.e. $F$ is differentiable on $[a, b] - E_{1}$ where $E_{1}$ is a subset of $[a, b]$ of measure zero. Similarly since $f$ is Riemann integrable on $[a, b]$, it follows that $f$ is continuous almost everywhere on $[a, b]$ so that there is a subset $E_{2}$ of $[a, b]$ of measure zero such that $f$ is continuous on $[a, b] - E_{2}$.
Since $E = E_{1} \cup E_{2}$ is also of measure zero, it follows that $F'(x) = f(x)$ for all $x \in [a, b] - E$. Thus the relation $F'(x) = f(x)$ holds almost everywhere in $[a, b]$.
Also note that $E_{1} \subseteq E_{2}$. I suspect there are examples where $E_{1} \subset E_{2}$ but I don't have an example right now with me.
