Suppose that $\mu$ is a measure and sigma-algebra $\Sigma$ over $X$. Suppose that $0<\mu(A)<\infty$ and for every $E\in \Sigma$ and $0<\mu(E)$ then $\mu(A\Delta E)=0$.
If we consider $L^p$ space over this domain, can we conclude that any function is equal to multiplication of indicator function of $A$ a.e.?
It's clear that $\mu(X\backslash A)=0$. So outside the measure of outside of $A$ is zero. I want to show that for any $f\in L^p$, there exists constant $c$ such that $f=c$ a.e. over $A$.
This is true. Hint: the hypothesis tells you that If $E$ is any measurable subset of $A$ then $\mu (E)=0$ or $\mu (A\setminus E)=0$. Take $x \in \mathbb R$ and let $E=A \cap f^{-1} (-\infty, x]$. We see that either $ f\leq x$ almost everywhere on $A$ or $ f> x$ almost everywhere on $A$. Let $c$ be the supremum of all $x$ such that $f \leq x$ almost everywhere. Can you show that $f=c$ almost everywhere?