What is $\int_{2}^{2}\frac{dx}{x-2}$? Evaluate the integral: $$\displaystyle\int_{2}^{2}\frac{dx}{x-2}.$$
1)When does $\displaystyle\int_a^a f(x)dx=0$? Always?
2)Does $\displaystyle\int_a^a$ means area between $(a,a)=\emptyset$?
3) Do we care end points? $\displaystyle\int_a^b$ means area between $(a,b)$ or $[a,b]$ or whatever?
4) I look to the Lebesgue integration, but not much. If we drop countable points, it is not matter. (I think). or at least finite points. It wont change the integral.
But if we think the definition of Riemann sum and partitions: it cares the end points $a\leq \cdots \leq b$.
But if we think the improper integral: It does not care the improper points and they are sometimes/generally end points.
I have a lot of questions in same manner in my mind. 
Thanks for any comment, idea or answer.
 A: The integral $\int_a^bf(x)\,dx$ is defined when $[a,b]$ is included in the domain of definition of $f$, and if so it is $=0$ when $a=b$.
When $(a,b)$ is included in the in the domain of definition, but $[a,b]$ is not, the integral $\int_a^bf(x)\,dx$ is defined as a limit.
In your case, I'd say that the symbol is not defined since there's no open interval over which you can take the limit of.
More or less like trying to define $\int_{-2}^{-1}\log(x)\,dx$.
A: *

*The integral $\int_a^a f(x)\,dx=0$ whenever $f(x)$ is defined at $a$, whether the integral is Riemann or Lebesgue (with Lebesgue measure). In the Riemann case, it is because every partition consists of rectangles with zero width. In the Lebesgue case, it is because the set $\{a\}$ has measure zero. In the case $f(x)=\frac{1}{x-2}$ and $a=2$, $f(2)$ is undefined, so the integral $\int_2^2 f(x)\,dx$ is also undefined.

*$\int_a^a\,dx$ means the Lebesgue measure of the set $[a,a]=\{a\}$, which is zero. There is some ambiguity here. It might also mean the Lebesgue measure of the set $(a,a)=\emptyset$, which is also zero. It's not a problem in this case, but it's better to write the unambiguous $\int_{[a,a]}\,dx$ or $\int_{(a,a)}\,dx.$

*The Lebesgue integral over an interval does not care about the endpoints. This is because the endpoints have measure zero. See point 2.

*Not sure what your question is, but yes, we can always drop countably many points. This is because a countable set always has Lebesgue measure zero. We can even drop uncountably many points if that uncountable set has Lebesgue measure zero, like the Cantor set.

