Is "integrability" equivalent to "having antiderivative"? I am wondering if "a function $f(x)$ is integrable on a domain $D$" this proposition is equivalent to "$f(x)$ has antiderivative on domain $D$". If it is not the case, give me a counter example. Thank you.
There is an even simpler example.  Take, e.g., 

$$
f(x) = \begin{cases}
  1 &amp; x \geq 0 \\
 -1 &amp; x &lt; 0
\end{cases}
$$

This function is Riemann-integrable on $[-1, 1]$, but it cannot be the derivative of any function because derivatives cannot have jump discontinuities.  (See, e.g., this question.)
No, they are two different concepts.
If a function y=f(x) has an antiderivative, it simply means that there exists a function, say F(x), whose derivative F'(x) = f(x). We say f is integrable on the interval [a,b] if the limit of its Riemann sums over this closed interval exits and is equal to some finite value.

Let's consider two different conditions:
1) when f is continuous on [a,b]
   when f is continuous, by the Fundamental Theorem of Calculus, it has antiderivative F(x)=∫xaf(t)dt, and F(x) is also continuous on [a,b].  Also, due to the property of a continuous function, f is integrable on [a, b]. In this case, having an antiderivative and integrable can be interpreted as the same meaning.

2) when f is not continuous on [a,b]
   we need to make this clear first. For a continuous and differentiable function, its derivative need not be continuous, that is, a non-continuous function may have an antiderivative; also, for a function to be integrable, it need not be continuous. And we discuss two questions:
   (a) Does the existence of an antiderivative guarantee that a function is integrable?
       No. As we know, a non-continuous function may have an antiderivative. Thus, the function may not be integrable. That is, although there is an explicitly defined antiderivative F(x) (possibly not, but just expressed as an integral using the Fundamental Theorem of Calculus), the limit of the Riemann sums may not exit or may not equal to some finite value. However, we limit our discussion here to any type of continuous functions, not necessarily bounded on the closed interval [a,b]. For bounded functions, we can expect something else. 
   (b) Does an integrable function guarantee the existence of an antiderivative?
       No. We can use the same reasoning as the previous argument. That is, a non-continuous function may be integrable. Thus, the function may not have an antiderivative. A famous example would be the Thomae's function. Which has been mentioned above.