On integral of simple function and representation

[psie]

TL;DR Summary

I'm trying to work out how the integral of a simple function is independent of the representation chosen for the simple function.

Definition 4.1. Let ##(X,\mathcal{A},\mu)## be a measure space. If ##\phi:X\to [0,\infty)## is a positive simple function, given by $$\phi=\sum_{i=1}^N c_i\chi_{E_i}$$ where ##c_i\geq 0## and ##E_i\in\mathcal{A}##, then the integral of ##\phi## with respect to ##\mu## is $$\int\phi \ d\mu=\sum_{i=1}^N c_i\mu(E_i).$$

I wonder, how does one show that the integral is independent of the representation of the simple function? Suppose $$\phi=\sum_{i=1}^N c_i\chi_{E_i}=\sum_{i=1}^M b_i\chi_{F_i}.$$ How does it follow then that $$\sum_{i=1}^N c_i\mu(E_i)=\sum_{i=1}^M b_i\mu(F_i)?$$

I have discussed this problem with someone else and they've told me that we first need to show that the integral of the simple function is not changed when we make the sets disjoint (by sets I mean those appearing in the simple function). With a bit of work, I think I've managed to show this and I can share some of that work if anyone's interested. However, how does it then follow that given two representations, we have that they correspond to the same integral? You are allowed to assume that the sets in the representations are disjoint, since this is what I've been able to show.

 

[anuttarasammyak]

May I interpret that
[tex]\mu(E_i)=\int \chi _{Ei} d\mu[/tex] ? Then the definition seems obvious.

 

[psie]

May I interpret that
[tex]\mu(E_i)=\int \chi _{Ei} d\mu[/tex] ? Then the definition seems obvious.

Indeed, this is true from the definition, but I don't see how given ##\phi=\sum_{i=1}^N c_i\chi_{E_i}=\sum_{i=1}^M b_i\chi_{F_i}##, we have ##\sum_{i=1}^N c_i\mu(E_i)=\sum_{i=1}^M b_i\mu(F_i)##. How would you show this?

 

[anuttarasammyak]

[tex]\phi=\sum_{i=1}^N c_i \chi_{E_i}=\sum_{i=1}^M b_i \chi_{F_i}[/tex]
[tex]\int d\mu \ \phi=\int d\mu\sum_{i=1}^N c_i \chi_{E_i}=\int d\mu\sum_{i=1}^M b_i \chi_{F_i}[/tex]
[tex]\int d\mu \ \phi=\sum_{i=1}^N c_i \mu(E_i)=\sum_{i=1}^M b_i \mu (F_i)[/tex]

 

[pasmith]

If two functions are equal, then their integrals must be equal. This follows from the fundamental nature of equality: equal objects may be freely substituted for each other.

Your question reduces to: how do you show that [itex]\sum c_i \chi_{E_i} = \sum b_i \chi_{F_i}[/itex]? The answer is: as you would for any other function. Two functions are equal iff they have the same domain and codomain and do the same thing to each element of their domain.

I think it is also implicit that the [itex]E_i[/itex] are disjoint, which means that if [itex]\sum c_i \chi_{e_i} =\sum b_i \chi_{F_i}[/itex] then for each [itex]i[/itex] there exists a unique [itex]j[/itex] such that [itex]c_i = b_j[/itex] and [itex]E_i = F_j[/itex].

 

[psie]

Thanks for the replies.

I found this proof on ProofWiki, which is quite self-contained.

 

[Office_Shredder]

I think everyone else in this thread is missing the main point, which is that the integral is not defined or proven to be consistent at this point in a standard analysis course - in fact this result is necessary to be able to define what an integral is