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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!

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    How do you mean unbounded? You could just redefine the function in all "bad" points if they remain to be a null set. Note that a countable set is of measure zero, but it can be way large. For example the Cantor set.2011-10-30
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    A function cannot be unbounded at a point. Whenever you think about a function, like say $\tan(x)$ which is not bounded "at" $\frac{\pi}{2}$, it is actually not bounded on any interval containing or touching $\frac{\pi}{2}$.2011-10-30
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    Maybe by terminology is poor. My apologies. So let me try an example: $\int_{-a}^a \frac{dx}{x^2}$ exists, but --- at least formally --- is not the same as $2\int_0^a \frac{dx}{x^2}$ because the latter integral doesn't exist. The difference between the two expressions is the origin, a set of measure zero, at the endpoint of the second integral.2011-10-30
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    What do you mean by "simply connectedness fails at the origin"? You mean at "the endpoints a=b"? Can you give me an example?2011-10-30
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    I'd rather say that the first does not exist, and the latter does. However both do exist when you use appropriate definitions.2011-10-30
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    @savick01: But the reason the 2nd doesn't exist is the singularity at the origin. So my question is why when integrating *through* this set of measure zero there's no problem, but *ending* the integral there it becomes ill-defined. On other words, why doesn't the "free pass" of measure zero apply at the endpoints?2011-10-30
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    @savick01: re: simply connected: I mean that Stokes' theorem only holds in a simply connected region -- even if the offending point is a set of measure zero. The simplest example is probably $\frac{\hat{\phi}}{r} = \frac{-y\hat{i}+x\hat{j}}{\,\,x^2+y^2}$. The circulation exists for any closed curve around the origin, but since the curl isn't defined there, Stokes fails.2011-10-30
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    It is very strange what you are writing about singularities at the origin and somewhere else. When I studied analysis, it was otherwise. What kind of integral are we talking about? Riemann / Leibniz-Newton?2011-10-30
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    @Jonas, the way I read your comment, it presents the Cantor set as an example of a countable set. Maybe you didn't mean it to be read that way. Anyway, the Cantor set is not countable.2011-10-30
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    @GerryMyerson No, I mean to say that measure zero sets can be way larger than countable sets.2011-10-31

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