TITLE: A mathematical fallacy concerning the integrability of a function and cancellation
QUESTION [12 upvotes]: I am reading the Florida Mu Alpha Theta Sponsors Guide. Page 43 is a list of clarifications and disputes commonly made, and their resolutions. One of their clarifications is this: 

A function which is not integrable on an interval A is not integrable on any interval B, where B contains A. I.e. no “the negative signs cancel” arguments.

What is this argument they're referring to?

REPLY [4 votes]: The rule that $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx $ should hold, and its generalization to all partitions of an interval into a finite number of closed sub-intervals.  This means $f$ should have an integral on all closed subintervals, if it is to be integrable.   
For the Riemann integral and its generalizations (Lebesgue, Stieltjes etc) the partition property is a consequence of $\int_a^b f $ existing.  Thus the word "should" can be replaced by "must" in the first paragraph.
Cancellation of singularities happens in some regularization schemes for divergent or improper integrals, which is a different matter that does not affect integrability.