# Thread: pulling out constants with the lebesgue integral

1. ## pulling out constants with the lebesgue integral

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?

2. 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!).

3. Originally Posted by HallsofIvy
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!).
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.

4. Nice book Rudin's!
You must distinguish 2 cases.

1. If $f$ is diferent form 0 in a subset F of E of positive measure, then it follows that $\int_E f\neq 0$ because $\mu\{x\in E:f(x)>1/n\}>0$ for some $n$. Hence $\infty f=\infty$ in F and hence $\int_E\infty f\geq \int_F\infty f=\infty\mu (F)=\infty=\infty\int_E f.$. Thus the inequality becomes an equality.

1 If $f$ equals 0 almost everyhere in $E$, then we can take $F$ with $\mu(E\setminus F)=0$ and $f=0$ on $F$. Therefore $\int_E \infty f=\int_F \infty f=\int_F 0=0=\infty 0=\infty\int_F f=\infty\int_E f$.