If $f(x)$ is discontinuous at $x=0$, can $\int_{-1}^1 f(x)dx$ exist. I am  interested in the reasoning.
All help is appreciated
 A: In general, when $f$ and $g$ differ in (at most) a finite amount of points (in the interval $[a,b]$), we always have
$$
\int _a^bf(x)\,dx=\int_a^b g(x)\,dx
$$
This follows from the definition using sums over (infima and suprema of) a division of the interval $[a,b]$ in smaller intervals and the fact that we can make arbitrarily small intervals around the points where the functions are different.
A: The integral is highly insensitive to small changes in the function. In particular, the function $f(x)=0$ for $x\ne 0$ and $f(0)=16$, as far as the integral is concerned, is the same as the constantly $0$ function. In particular, both are integrable and the integral is $0$. 
A: Example: $f(x)=0$ for $x<0$ and $f(x)=1$ for $x>0$. Another example: $f(x)=\ln|x|$.
A: Given the way you ask your question, I assume that you are unfamiliar with the definition of the Riemann Integral. Let the upper Riemann-sum $U$ of a function $f$ on a partition $P = \{x_0, x_1, x_2 \cdots x_n\}$ be defined as $$\sum_{i = 1}^n \Delta x_i\cdot I^*$$ where $\Delta x_i = x_i - x_{i-1}$ and $I^*$ is the greatest value of $f$ on $[x_{i-1}, x_i]$. The lower Riemann-sum $L(f, P)$ is defined in the obvious way.
Definition 1: A function is Riemann Integrable if there exists a partition $P$ so that for every $\epsilon > 0$ we have that $U(f, P) - L(f, P) < \epsilon$
Consider this slightly edited function provided by Daniel Fischer as a comment to your question: $$f(x) = \begin{cases} -1 &,-1 \le x < 0 \\ 0 &, x = 0\\ 1 &,1 \ge x > 0 \end{cases}$$
Let $P = \{-1, 0-\epsilon, 0+\epsilon, 1\}$. We have $U(f,P) = (-1) \cdot(1-\epsilon) + 1\cdot2\epsilon + 1\cdot (1-\epsilon) = -1 + \epsilon + 2\epsilon + 1 - \epsilon = 2\epsilon$ and $L(f,P) = (-1) \cdot (1-\epsilon) + (-1)\cdot 2\epsilon + 1\cdot(1-\epsilon) = -1 + \epsilon - 2\epsilon + 1 - \epsilon = -2\epsilon$. So we have that $U(f,P) - L(f,P) = 4\epsilon$. The function is still integrable, although our value is greater than $\epsilon$. Can you tell why? Can you see how you might configure the values of the partition so that the result is prettier? 
