Does every continuous function has an anti-derivative? How come any continuous function has an anti-derivative?
I am interested in knowing whether for every continuous function f on [a,b] there exists some F whose derivatives on [a,b] equal to f, but domain not necessarily f’s. An example would be f(x) = 0 for x in [-1, 1], and F(x)=1 for all real x.
I have seen a lot of supposed proofs claim the integral function in the First Fundamental Theorem of Calculus is an anti-derivative of any continuous f on [a,b]. That is clearly wrong,since we all know it is only differentiable on (a, b).
Suppose f is continuous and defined only on [a,b], then F (the integral function in the First Fundamental Theorem of Calculus) is consequently defined only on [a,b] and clearly not differentiable on a nor b since F is simply not defined for values outside the domain of f.
But the fact that any continuous function has an anti-derivative is frequently used in proofs for other important theorems like the substitution rule, seeing from Wikipedia, I guess there might be some other proof, what is it?

 A: It is not true that every continuous function has an antiderivative. Its domain matters. No continuous function $f:[a,b]\to\mathbb R$ has an antiderivative, since such an antiderivative would have to be differentiable on a non-open set, which is impossible by definition. The derivative is only defined at interior points of the domain.
However, if we allow one-sided differentiability, which some authors do, then continuous functions on closed intervals do have an antiderivative, given by the integral function $I_{f,a}$, which integrates starting at $a$. This is differentiable (from the right) at $a$, since
$$\begin{align}
\frac{I_{f,a}(x)-I_{f,a}(a)}{x-a}&=\frac{1}{x-a}\left(\int_a^x f(t)\mathrm dt-\int_a^{a}f(t)\mathrm dt\right)\\
&=\frac{1}{x-a}\int_a^x f(t)\mathrm dt\\
&=f(a)-f(a)+\frac{1}{x-a}\int_a^x f(t)\mathrm dt\\
&=f(a)+\frac{1}{x-a}\int_a^x f(t)-f(a)\mathrm dt.
\end{align}$$
The absolute value of the term with the integral in the end is bounded by $\frac{\vert x-a\vert\sup_{t\in[a,x]}\vert f(t)-f(a)\vert}{\vert x-a\vert}=\sup_t\vert f(t)-f(a)\vert$, which, by continuity of $f$, goes to $0$ as $x\to a$, and then the difference quotient above goes to $f(a)$. By a similar argument it is differentiable (from the left) at $b$.
Note that this is essentially the proof of the FTC, with no alterations.
If you don't want to go the route of one-sided differentiability, then the theorems like integration by parts need more careful consideration of the boundary points of the domain. You would need to show that the functions appearing in the theorems can be continuously extended to the boundaries, and then by uniqueness of the continuous extension you could claim that the equations hold for those extensions, too. For instance,
$$\int_\alpha^\beta f'(t)g(t)\mathrm dt=f(\beta)g(\beta)-f(\alpha)g(\beta)-\int_\alpha^\beta f(t)g'(t)\mathrm dt$$
holds for all $a<\alpha<\beta<b$, and if the two sides of the equation as functions of $\alpha$ and $\beta$ can be continuously extended to $(a,b)$ (to be read as a tuple, not an interval), then the equations also hold for the continuous extensions, which we write by replacing the Greek letters by their Latin counterparts.
