set of measure zero at end of closed interval I'm sure this is elementary, but:
If a 1-D function is bounded everywhere except at a point P (measure zero), then the integral exists.  So why should it matter if P is at the end of the integration interval?  
Presumably the same is true of a line integral in, say, R3 --- which would seem to imply that the path for a closed line integral can include a countable number of discontinuities.   But what happens if the closed integral is calculated in such a way that the endpoints a=b are where the function is unbounded?    
Taking this one more step, the closed line integral, by Stokes' theorem, can be written as a surface integral of the curl.  As long as the regions is simply connected.  But if, say, simply connectedness fails at the origin, why should it matter -- isn't that just a set of measure zero for the surface integral?
Clearly, I'm confused and/or missing something basic.
thanks!
 A: It does matter because when you calculate the integral of $\int_0^1 x^{-2} dx$, you actually calculate a different KIND of integral than the integral say $\int_0^1 x dx$.
If you go back to definitions: a 'normal' Riemann integral is a limit of Riemann sums. When you want to define $\int_0^1 x^{-2} dx$ you can't use this definition, because any sum can be arbitrarily huge - it is a matter of taking an appropriate midpoint near $0$, which generates arbitrarily big value. So you say: I can calculate a limit of integrals $\lim_{\varepsilon \to 0}\int_\varepsilon^1 x^{-2} dx$. But look: it is kinda trick, that you take the limit. If you want, you can say: I want the number: $\lim_{\varepsilon \to 0} \int_{-1}^{-\varepsilon} x^{-2} dx + \int_\varepsilon^1 x^{-2} dx$, to be the value of the integral $\int_{-1}^1 x^{-2} dx$.
That really shows weakness of the Riemann integral, so people usually use different definitions (like Lebesgue integral), to be able to calculate the integral of the characteristic function of $\mathbb{Q}$ or things like that. They prove that it is equal to traditional Riemann integral if Riemann integral is defined. However there are some cases where Riemann integral in the second version (with $\lim$) is defined and Lebesgue integral is not: that is because for example $\int_1^\infty{}\frac{sin(x)}{x}$ for Riemann is the limit when you go to infinity and Lebesgue looks at the whole halfline and he sees infinitely big surface (I mean "the surface below the graph") above zero and below zero and he does know how much $\infty - \infty$ is.
At the beginning you say bounded and then you switch to discontinuous, I don't get it. You mean that the path is not continuous or the function which is integrated along the path? I haven't heard about discontinuous path (see: path), but integrals on $\mathbb{R}$ and line integrals have indeed a lot in common.
Simple connectedness - it is a term from topology not measure theory and in topology single points sometimes really matter. For a topologist a circle without one point is an interval which is of course completely different from a circle (it is simply connected for example, but you can also notice that if you take one point out of a circle it is still connected, what is not true in case of interior points of an interval). For a topologist a disk without one point is in some way more similar to the circle than to the whole disk. EDIT: So topology sees no difference if in your formula there is a singularity at zero or at some disk (with positive measure) around it. END
Going back to simple connectedness (s.c): taking one point from an open disk spoils its s.c. And it does matter if you - for example - want to say that particular line integrals are equal, because the endpoints are equal - if you have s.c, you can use a homotopy between those paths to prove such a statement and you have a problem, when there is no homotopy (because of no s.c.). In that case an infinite value at one single point may spoil everything. Well, such is life.
There are actually no theorems in my answer, but I hope it helps a bit.
P.S. If you calculate the 'limit' Riemann integral and normal Riemann integral exists, they are equal, so you can use only the 'limit' one, but it is somehow pointless.
A: EDIT: as pointed out in the comments, even the principal value interpretation doesn't work for this particular function. I had in mind the situation where a principal value exists because positive and negative contributions cancel out, e.g., for ${\rm p.v.}\int_{-1}^1(1/x)\,dx=0$. So much of what follows is wrong and/or not relevant to the question at hand. 
I would say that $\int_{-1}^1(1/x^2)\,dx$ doesn't exist, and neither does $\int_0^1(1/x^2)\,dx$. The first integral differs from the second one in that there is something called the "principal value" defined by $${\rm p.v.}\int_{-1}^1{1\over x^2}\,dx=\lim_{a\to0}\left(\int_{-1}^{-a}{1\over x^2}\,dx+\int_a^1{1\over x^2}\,dx\right)$$ and the principal value does exist, while no principal value exists for the second integral. But the principal value isn't the integral; it's a generalization of the integral. 
