I am studying measure theory on my own and there is something about the convention that $0 \times \infty = 0$ that I can's seem to get my head around. I've read various threads now on this topic and it seems that everyone agrees that it cannot be more than a convention, and thus we can't use it to prove any results (see e.g. Measure theory convention that $\infty \cdot 0 = 0$)
However, in most sources this convention is introduced right after the definition of the integral of a simple function.
Def: Simple function
Let $(\Omega,\varphi,\mu)$ be a measure space. Then a function $f: \Omega \to \mathbb{R}$ is a simple function if $f=\sum_{j=1}^{n} c_{j} \mathbb{1}_{E_{j}} $ s.t. $c_{j} \in \mathbb{R}$,$E_{j} \in \varphi$,$\sum_{j=1}^{n} E_{j}=\Omega$ and $E_{j} \cap E_{k}=\emptyset$ if $j \neq k$.
Def: Integral of simple function
Let f be a non-negative simple function, then we define its integral as
$I(f)=\sum_{j=1}^{n} c_{j} \mu(E_{j})$.
With this definition we can run into problems if $c_{j}=0$ and $\mu(E_{j})=\infty$ for some $j$. Is there a way to show that the value of the integral does not depend on the convention $0 \times \infty = 0$. I know that it is possible to prove that the value of the integral does not depend on the particular representation of a simple function, so my idea is to circumvent the problem by using a different representation of a simple function if we run into the case described above. The problem is that I can't find a way how I could prove that for a general function $f$.
Appreciate any help on this. Thanks very much!
I don't think what you want can be achieved. Say $\Omega$ is the real line with its Borel $\sigma$-algebra and let $\mu $ be the Dirac measure on $0$. Then any partition of $\Omega$ will have a member with infinite measure, so if $f$ is the zero function you will never be able to avoid the situation $c_j=0$ and $\mu(E_j)=\infty$ for some $j$.