# Definition of the Lebesgue integral in terms of simple functions with finite measure support

Let $(X,\Sigma,\mu)$ be a measure space. A function of the form $$\phi(x) = \sum_{i=1}^{n} c_i \mathbf{1}_{E_i}$$ where $c_i \in \mathbb{R}$ and $E_i \in \Sigma$ is called a simple function. If $c_i \geq 0$ for all $i$, the Lebesgue integral of $\phi$ is $$\int \phi d\mu = \sum_{i=1}^{n} c_i \mu(E_i).$$ Let $f$ be a non-negative measurable function. If $X$ is $\sigma$-finite, we have \begin{align}\tag{$\ast$} \sup\left\{ \int \phi d\mu : \phi \text{ simple, } 0 \leq \phi \leq f \right\} = \sup\left\{ \int \phi d\mu : \phi \text{ simple, } 0 \leq \phi \leq f, \, \mu(\{x: \phi(x) \neq 0\}) < \infty \right\} \end{align} and their common value is $\int f d\mu$, the Lebesgue integral of $f$.

To see that $(\ast)$ holds, it suffices to consider sets $X_1, X_2, \ldots, \in \Sigma$ with $X = \cup_{i=1}^{\infty} X_i$ and $\mu(X_i) < \infty$ and note that $$\lim_{N \rightarrow \infty} \sum_{i=1}^{n} c_i \mu(E_i \cap \cup_{i=1}^{N} X_i ) = \sum_{i=1}^{n} c_i \mu(E_i)$$ because, for each $i$, the sequence of sets $\{ E_i \cap \cup_{i=1}^{N} X_i \}_{N=1}^{\infty}$ is increasing and $$\bigcup_{N=1}^{\infty} (E_i \cap \cup_{i=1}^{N} X_i) = E_i.$$

Question. Does $(\ast)$ hold if we relax or drop the hypothesis that $X$ is $\sigma$-finite?

Try taking $X=[0,1], \Sigma = \mathcal P \left( {[0,1]} \right), \mu(A)=0$ if $A$ is finite or countable, and $\mu(A)=\infty$ if $A$ is uncountable. Then $\sigma$-finiteness does not hold on this measure space.
Consider $f(x)=1.$ Notice $f$ is $\mu$-measurable. Consider for $(*)$, $$\sup \left\{ \int_{X} \phi d\mu:\phi \space \text{simple}, 0 \leq \phi \leq 1\ \right\} =\int_{X} 1 d\mu=1 \cdot \infty=\infty$$ but for the right hand side of $(*)$, $$\sup \left\{ \int_{X} \phi d\mu:\phi \space \text{simple}, 0 \leq \phi \leq 1\ , \mu\{x:\phi(x) \ne 0\} < \infty \right\}=0$$
since for all such functions $\phi$ in this case, $$\mu\{x:\phi(x) \ne 0\}< \infty$$ if and only if $$\mu\{x:\phi(x) \ne 0\}=0$$ so that $$\int_{X} \phi d\mu \leq 1 \cdot 0=0$$ for all such $\phi$.
Let $X$ be any non-empty set, $\Sigma = \{\emptyset, X\}$, $\mu(\emptyset)=0$, and $\mu(X)=\infty$. Note the only simple function $\phi$ with $\mu(\{x:\phi(x) \neq 0\}) < \infty$ is $\phi \equiv 0$. So if $f \equiv 1$ we have $$\sup \left\{ \int \phi d\mu:\phi \text{ simple, } 0 \leq \phi \leq f\ \right\} =\int 1 d\mu=1 \cdot \infty = \infty$$ and $$\sup\left\{ \int \phi d\mu : \phi \text{ simple, } 0 \leq \phi \leq f, \, \mu(\{x: \phi(x) \neq 0\}) < \infty \right\} = \int 0 d\mu = 0.$$