Is this integral defined? $\int_0^0\frac 1x\:dx$ $\displaystyle\int_0^0\frac 1x\:dx$
Is this integral defined? If it define, what is the value?
 A: It is defined, and has value $0$: write $I=\int_{0}^{0}\frac{dx}{x}=\int_{-1}^{1}\frac{\chi_{\{0\}}dx}{x}$, where $\chi_{A}$ is the indicator function on set A, $\chi_{A}(x)=1$ if $x$ belongs to $A$, and is $0$ otherwise. Now $\{0\}$ has Lebesgue measure $0$, and outside of $\{0\},\frac{\chi_{\{0\}}}{x}=0$. Hence $I=\int_{-1}^{1}0=0$. Thus $I$ is defined, at least in the sense of Lebesgue. 
A: I will take the case of Riemann Integral because the Lebesgue part is nicely done by Noe Blassel.
The definition of Riemann $\int_{a}^{b}f(x)\,dx$ has a pre-condition that $f$ is bounded on $[a, b]$. This itself requires that $f$ must be defined for all points in $[a, b]$. However we are still able to talk about Riemann integrals like $\int_{0}^{1}\sin(1/x)\,dx$. How??
Note that Riemann integrals have a nice property that the value of integral $\int_{a}^{b}f(x)\,dx$ does not depend on values of $f$ on a finite number points in $[a, b]$. Thus we simply define the function $f(x) = \sin (1/x)$ at $x = 0$ in any arbitrary manner say $f(0) = k$ and the function remains bounded in interval $[0, 1]$ and since its only discontinuity is at $x = 0$ the function is Riemann integrable and the integral makes sense.
Next we come to the function $f(x) = 1/x$ on interval $[0, 0]$. Here $f$ is not defined on the interval under consideration. Hence it does not make sense to talk about its integral unless we define it on that interval. So we can define it in any manner by setting $f(0) = k$ and then $f$ is bounded on $[0, 0]$ and the integral is equal to $0$ (if we use upper/lower sums both are $0$).
Note that the argument of the previous case applies to any function on any interval of length $0$ and we expect the integral to be $0$ (via upper / lower sums approach). Hence while studying Riemann integrals we add the following extra definitions:
1) If $f$ is any function then we define $\int_{a}^{a}f(x)\,dx = 0$.
2) If $f$ is Riemann integrable on $[a, b]$ (with $a < b$) then we define the symbol $\int_{b}^{a}f(x)\,dx$ as $-\int_{a}^{b}f(x)\,dx$
Together these two definitions help us to write the following theorem:
Theorem: If $a, b, c$ are any points lying in some closed interval $I$ on which $f$ is Riemann integrable then $$\int_{a}^{b}f(x)\,dx + \int_{b}^{c}f(x)\,dx = \int_{a}^{c}f(x)\,dx$$ irrespective of any order relations between $a, b, c$.
