Edit: My post might be lacking context. The question was posed because the book has only defined the Lebesgue integral of functions on a measurable set $E$ where $m(E)<\infty.$ The definition below asks, if given a measurable $E$ whose measure can be finite or infinite, is it really properly defined?
This is a problem from Royden and Fitzpatrick's Real Analysis. Can anybody check if my proof is correct?
Suppose $f$ is a bounded and measurable function on E with a finite support, we can define its integral over $E$ by $$\int_E f = \int_{E{_0}} f$$ where $E_0$ has finite measure and $f\equiv 0$ on $E \sim E_0$.
My attempt at proof. We wish to show that the definition above is independent of the choice of finite measure set $E_0$ where $f$ vanishes outside it.
By hypothesis, $f$ has finite support, denote the support of $f$ by $E'=\{x\in E: f(x) \neq 0\}$. Let $E_0$ and $E_1$ be subsets of $E$ of finite measure where $f$ vanishes outside it. By the additivity over domains property of integration, since $E' \subseteq E_0$, and $E' \subseteq E_1$ (so $E'$ has finite measure as well), then we can write: $$\int_{E_0}f=\int_{E_0 \sim E'}f +\int_{E'}f, $$ and $$\int_{E_1}f=\int_{E_1 \sim E'}f +\int_{E'}f.$$ Therefore: $$\int_{E_0}f-\int_{E_1}f =\int_{E_0 \sim E'}f – \int_{E_1 \sim E'}f.$$ But, $E_0 \sim E'=\{x\in E_0 :f(x)=0 \}$ and $E_1 \sim E'=\{x\in E_1 :f(x)=0 \}$, so $f\equiv 0$ in both of these sets, hence, the integral of $f$ over these sets is $0$. Therefore: $$\int_{E_0}f=\int_{E_1}f.$$
Hence, the definition above is well defined since $E_0$ and $E_1$ are arbitrary.
Best Answer
From what I understand, your prerequisites are : you know how to define the integral of functions on measured spaces with finite measure ; you know the additivity of the integral with respect to the domain ; you know that the integral of a zero function is zero.
On the other hand, I don't really understand the point, since if you want to define the integral of a measurable function $f:E\to \mathbb{R}$ with finite support, then since $f$ is measurable the support $E'$ is measurable, and has finite measure by hypothesis, so you may directly define $$\int_E f = \int_{E'} f.$$
PS : you should probably not use $E \sim F$ for set complementation, it's very likely confusing for most people (me included).