- #36
mathwonk
Science Advisor
Homework Helper
- 11,801
- 2,051
After some thought, I can imagine cases where one would like to distinguish between integrals that seem to have infinite value, as opposed to those to which no value can be assigned, finite or infinite. I can imagine proceeding as follows: Separate the function f into the difference of two non negative functions f+ and f-, by separating the parts of the graph that are above the x-axis from the part below it. I.e. f+(x) = max{f(x), 0}, and f-(x) = max{-f(x),0}.
Then say a non negative function f is integrable if it is the pointwise limit from below a.e. by non negative step functions, and if so, that its integral is the sup of the integrals of those step functions. Then the integral is finite if the integrals of those step functions can be taken to be bounded above. Then, one can define the integral of f by subtracting the integrals of f+ and f-, provided at least one of f+ or f-, has finite integral.
So one loses in this theory some familiar properties, such as the ability to add integrals, unless they are of the same sign, or at least one is finite. Fubini's theorem is also somewhat more complicated. Even if one deals entirely with positive valued functions, the key question to ask, regardless of the terminology, seems to be whether the integral is finite or not. When functions have both positive and negative values, here is a simple example from Fleming showing how Fubini can go wrong: the function f(x,y) = (1/y).cos(x), on the square with both x and y in the interval [0,π]. If y is constant the x -integrals are all zero, but if x is constant, the y integrals are almost all infinite, so the double integral does not exist even though one of the repeated ones does exist and is finite.
Then in this more general terminology, a function f would fail to be integrable if either it is too rough, or both f+ and f- are too big. So the essential issues are the same.
Then say a non negative function f is integrable if it is the pointwise limit from below a.e. by non negative step functions, and if so, that its integral is the sup of the integrals of those step functions. Then the integral is finite if the integrals of those step functions can be taken to be bounded above. Then, one can define the integral of f by subtracting the integrals of f+ and f-, provided at least one of f+ or f-, has finite integral.
So one loses in this theory some familiar properties, such as the ability to add integrals, unless they are of the same sign, or at least one is finite. Fubini's theorem is also somewhat more complicated. Even if one deals entirely with positive valued functions, the key question to ask, regardless of the terminology, seems to be whether the integral is finite or not. When functions have both positive and negative values, here is a simple example from Fleming showing how Fubini can go wrong: the function f(x,y) = (1/y).cos(x), on the square with both x and y in the interval [0,π]. If y is constant the x -integrals are all zero, but if x is constant, the y integrals are almost all infinite, so the double integral does not exist even though one of the repeated ones does exist and is finite.
Then in this more general terminology, a function f would fail to be integrable if either it is too rough, or both f+ and f- are too big. So the essential issues are the same.
Last edited: