I had to solve this question for a group assignment, and my group is suspicious of my proof for this exercise. Can anyone help me confirm if this proof is correct?
Bartle's Elements of Integration - Exercise 4.S: Let $f: X \to \mathbb{R}\cup\{-\infty, \infty\} = \overline{\mathbb{R}}$ be a non-negative measurable function, where $X$ is a measure space, such that $\int f d \mu < \infty$. Then, for every $\varepsilon > 0$, there exists a measurable set $E$, with $\mu(E) < \infty$, such that $\int f d \mu \leq \int_E f d \mu + \varepsilon$.
So here is my proof: Let $\varepsilon > 0$. Then there is a simple function $\varphi$ such that $\int \varphi d \mu > \int f d \mu - \varepsilon$ and $\phi(x) \leq f(x)$ for each $x \in X$ (this follows from Bartle's definition of the integral). Let $\{a_1, \dots, a_n\}$ be the set of the distinct positive values $\varphi$ attains, and $E_i = \varphi^{-1}(a_i)$. Then we can write $\varphi = \sum_{i=1}^{n}a_i \chi_{E_i}$ - where $\chi_A$ is the characteristic function of $A$. For each $E_i$, $\mu(E_i) < \infty$, because $a_i \mu(E_i) \leq \int \phi d \mu \leq \int f d \mu < \infty $, and $a_i \neq 0$. Then $E = E_1 \cup \cdots \cup E_n$ is such that $\mu(E) < \infty$ and $\varphi \chi_E = \varphi$. From this, we have that $\varphi(x) = \varphi(x) \chi_E(x) \leq f(x) \chi_E(x)$. Thus: $$\int_Efd \mu = \int f \chi_E d \mu \geq \int \varphi d \mu > \int f d \mu - \varepsilon$$ And from this follows that $\int f d \mu < \int_E f d \mu + \varepsilon$.
Is this proof correct?