dense subalgebra in measurable functions set

Question

Take $\mathcal M_D$ the space of measurable functions from a compact set $D\subseteq \mathbb R_n$ to ℂ.

I'm wondering if a Stone-Weierstrass-like theorem holds in this space, with the convergence in measure.

In other words, suppose $\mathcal A \subseteq \mathcal M_D$ is a closed proper subalgebra with identity(constant function 1), also closed by conjugation. Is it true that there exists $E\subseteq D$ measurable set with positive measure such that all functions in $\mathcal A$ are constant (almost everywhere) on $E$?

A similar question was asked here for $L^p$ functions, where the answer is positive, but here we haven't a norm.

Answer 1

Take $D = [0,1]$ and let $\mathcal{A}$ be the set of all measurable functions satisfying $f(x) = f(1-x)$ almost everywhere. It satisfies your conditions but contains the function $x(1-x)$ which isn't constant on any set with more than two points.

(This illustrates that "separates points", in the sense of the previously asked question about $L^p$, is not the negation of "there is a set of positive measure where everything is constant".)

I might guess that under these assumptions, the following conclusion could be true: there is a complete sub-$\sigma$-algebra $\mathcal{F}$ of the Lebesgue measurable sets, such that $\mathcal{A}$ consists of all the $\mathcal{F}$-measurable functions. I'd try to apply the multiplicative system theorem https://math.stackexchange.com/a/47521/822 which for me is the "measurable" analogue of Stone-Weierstrass.

Answer 2

This is a partial answer for the case that your algebra is inverse closed, i.e., if $f,g$ are in the algebra and $g$ is invertible, then so is $\frac f g$.Consider the intersection of your algebra with $L^\infty$ and apply the fact about the latter which you refer to. Now use the fact that for any measurable $f$, the quotient $\dfrac f{1+|f|^2}$ is bounded and its sets of contancy coincide with those of $f$.