Density of simple functions in a generating $\pi$ system for $L^p$ Let $(E, \mathcal{E}, \mu)$ be a measure space and define the simple functions to be of the form
$$\sum_{k = 1}^{n} a_{k} \mathbf{1}_{A_{k}}$$
where $a_{k} \in \mathbb{R}$ and $A_{k} \in \mathcal{E}$ with $\mu(A_k) < \infty$. We know that these functions are dense in the Lebesgue space $L^{p}(E, \mathcal{E}, \mu)$ when $p \in [1, \infty)$. See for example Theorem 3.13 in Rudin, Real and Complex Analysis.
My question: does this still hold true when we restrict only to have $A_{k} \in I$, where $I$ is a $\pi$ system which generates $\mathcal{E}$, that is $\mathcal{E} = \sigma(I)$? Let's call such functions extra simple functions.
It would suffice then to show that the extra simple functions can approximate $\mathbf{1}_A$ where $A \in \mathcal{E}$ is arbitrary with $\mu(A) < \infty$.
 A: The answer is relatively easy for finite measure spaces. I am also going to assume that $E \in I$ for now. Write
$$ \Sigma(I) := \left\{ \sum_{i=1}^n a_i \chi_{A_i} \ | \ a_i \in \mathbb{R}, A_i \in I \right\},$$
$$ \Sigma_p(I) := \Sigma(I) \cap L^p.$$
Our goal is to show that $\Sigma_p(I) \subseteq L^p$ is dense. Apriori, we know that $\Sigma_p(\mathcal{E}) \subseteq L^p$ is dense. Let
$$\mathcal{E}_f := \{A \in \mathcal{E} \ | \ \mu(A) < \infty\}. $$
Assume this is non-empty, since otherwise we're done.
Remarks:

*

*This is not a $\sigma$-algebra unless we are dealing with a finite measure space. It is a $\pi$-system.


*We have that $\Sigma_p(\mathcal{E}_f)$ is also dense inside of $L^p$.
Let
$$ D_p(I) := \{A \in \mathcal{E} \ | \ \chi_A \in \overline{\Sigma_p(I)}\}.$$
The big idea (at least for me) is that there's a sort of duality going on here between functions and sets via the map $A \mapsto \chi_A$. So $D_p(I)$ is a sort of double dual of $I$ inside the $\sigma$-algbera. Since we're dealing with $L^p$, this analogy stops being useful since $I$ doesn't fit nicely inside of $L^p$.
In general:

*

*$D_p(I)$ is a $\pi$-system.

If our space is a finite measure space, then:

*

*$I \subseteq D_p(I)$, since $\chi_A \in L^p$ for all $A \in I$.


*$D_p(I)$ is a $\lambda$-system.
Hence for finite measure spaces, $D_p(I) = \mathcal{E} = \mathcal{E}_f$ and we get that $\chi_A \in \overline{\Sigma_p(I)}$ for every $A \in \mathcal{E}_f$ (and this duality analogy makes sense -- double duals generally lead to closures, and this is a sort of measure theoretic closure).
For general measure spaces:

*

*$I$ need not be inside $D_p(I)$ -- for example, take $I = \{(-\infty, a] \ | \ a \in \mathbb{R}\}$. This $\pi$-system generates the Borel $\sigma$-algebra, and it is not inside $D_p(I)$. So we can see the duality analogy dies at the easiest example.


*In the above example, $D_p(I)$ does contain a nice $\pi$-system which also generates the $\sigma$-algebra. So to maybe fix the analogy, we need to study $\sigma(D_p(I))$ and see how this relates to $\Sigma_p(I)$ (without writing it down, I'm guessing it doesn't relate well).


*$D_p(I)$ need not be a $\lambda$-system. See the above example.
Remarks:

*

*I'm guessing for $\sigma$-finite one can probably break things up inside finite measure sets and then use the finite measure case to get the result (there are some technical details that I wasn't able to immediately resolve).


*For the non-$\sigma$-finite case, there might be a trick to just reduce to the $\sigma$-finite, but it's unclear.
