Can there exist a function with discontinuity at Cantor's Set union Z? I know there can't exist function with discontinuities only at irrational points,since cantor set is also uncountable like irrational numbers,I thought that the answer is no. Also if yes can you give me the function explicitly?
 A: The characteristic function of the Cantor set $C$, given by
$$f(x) = \begin{cases}
1 & x \in C\\
0 & x \notin C,
\end{cases}
$$
is discontinuous at all points in $C$ and continuous everywhere else.  You can easily modify it to add discontinuities at every integer (although I'm not sure why you'd want to do that.)
A: $\chi_C$, the characteristic function of the Cantor set $C$, is discontinuous precisely at the points of $C$.
To see why: $C$ is closed. For every point $x \not \in C$ there is an open neighborhood $U$ of $x$ such that $U \cap C = \emptyset$. $\chi_C$ is constant on $U$. It follows that $\chi_C$ is continuous on $\mathbb R - C$. Discontinuity on $C$ follows similarly.
Now modify $\chi_C$ to be discontinuous on $\mathbb Z$ too.

As @bof points out in the comments, the set of points of discontinuity of a function $\mathbb R \to \mathbb R$ is an $F_\sigma$ set. The Cantor set is one such set. The set of irrational numbers is not, however, by the Baire category theorem.
A: The set of discontinuities of a function must be an $F_\sigma$ set--a countable union of closed sets, and every $F_\sigma$ set is the set of discontinuities of some function. One can readily show that if $C$ is the Cantor set, then $E=\Bbb Z\cup C$ is indeed $F_\sigma$, since it is closed.
As for a function that is discontinuous precisely on $E$, simply consider the characteristic function of the set $E$--that is, the function $\chi_E:\Bbb R\to\Bbb R$ given by $$\chi_E(x)=\begin{cases}1 & \text{if }x\in E\\0 & \text{otherwise.}\end{cases}$$
