Infinity is not even a number. I can't imagine why you would want to multiply a function by infinity (nor how you would do it!).
Let f be a positive, measurable function. Let c be a constant. let E be a measurable set.
I know that the integral of cf over E equals c times the integral of f over E when c is greater than or equal to 0 and less than infinity. I need to prove that this works even when c= infinity. I am a little confused because isn't infinity times a positive value infinity? I suppose 0 times infinity is 0. Does it suffice to just look at these two cases?
I am also sort of skeptical but it is an exercise in Rudin's Real and Complex Analysis text.
In the extended reals we define infinity, call it q, like this:
let a be in [0,q]
then
a+q=q+a=q for all real numbers a
aq=qa= q if a is nonzero and 0 if a is zero.
This way we preserve commutative/associative and distributive properties.
We have to be careful with the cancellation laws: we can't subtract or divide by q.
Nice book Rudin's!
You must distinguish 2 cases.
1. If is diferent form 0 in a subset F of E of positive measure, then it follows that because for some . Hence in F and hence . Thus the inequality becomes an equality.
1 If equals 0 almost everyhere in , then we can take with and on . Therefore .