Partition a divergent integral into finite values

Main Question or Discussion Point

Hi there,

I am reading an article, but I faced the following problem, and I am wondering if it is well known fact.

If the integral of a function on some interval is infinity, can we partition this interval into countable disjoint (in their interiors) subintervals such that the integral on each interval is 1/2 for example?

Likemath

mathman
Hi there,

I am reading an article, but I faced the following problem, and I am wondering if it is well known fact.

If the integral of a function on some interval is infinity, can we partition this interval into countable disjoint (in their interiors) subintervals such that the integral on each interval is 1/2 for example?

Likemath
It looks like it is doable. You could use any positive finite number. (I assume you mean the integral over each piece = 1/2, so that the total is the sum of an infinite number of 1/2's).

Thank you, but how can I convince my self that it is doable?

mfb
Mentor
Counterexample:
##f:[0,3] \to R##
##f(x)=\frac{1}{1-x}## for ##0\leq x<1##
##f(x)=-1## for ##1\leq x \leq 2##
##f(x)=\frac{1}{x-2}## for ##2 < x \leq 3##

The ugly part is the middle section: You cannot combine it with any interval of the other two sections, as this would diverge.

It is true for a large class of functions, however: if ##\int_a^x f(x') dx'## is well-defined for every x in your interval (a,b), it is continuous and you can find appropriate intervals with the intermediate value theorem. This directly gives a way to count them, too, of course.