Let $A$ be a complex algebra of bounded measurable functions on the measure space $(X,\mu)$ (case of $[0,1]$ with Lebesgue measure is enough for me) closed under conjugation. Assume that $A$ separates points, i.e. there is no non-trivial measurable partition of $X$ such that each function in $A$ is constant on (almost every) part. Is it true that $A$ is dense in $L^p(X,\mu)$ for $1\leq p < \infty$? 



Stone-Weierstrass analogue for $L^p$
Yes (I assume that the measure is finite). Here is a proof that uses the von Neumann bicommutant theorem (or rather [Kaplansky's density theorem](http://en.wikipedia.org/wiki/Kaplansky_density_theorem)).


See $A \subset L^\infty(X,\mu) \subset B(L^2(X,\mu))$ where $L^\infty$ acts on $L^2$ by pointwise multiplication. Then the assumption that $A$ separates points is exactly that the commutant of $A$ is $L^\infty(X,\mu)$, so that the bicommutant of $A$ is $L^\infty(X,\mu)$. Therefore, by Kaplansky's density theorem, any $f \in L^\infty$ with $\|f\|\_\infty \leq 1$ belongs to the strong operator topology closure of $\{g \in A, \|g\|\_\infty\leq 1\}$. Equivalently, there is a net $g\_\alpha \in A$ with $\|g\_\alpha\|\_\infty \leq 1$ such that, for every $\xi \in L^2$, $\|g\_\alpha \xi - f \xi\|\_2\to 1$. In particular (using that the constant function $1$ belongs to $L^2$), $\|g\_\alpha - f\|\_2 \to 0$. But by this implies that for every $1\leq p < \infty$, $\|g\_\alpha - f\|\_p \to 0$~: if $p<2$ this is because the $L^p \subset L^2$ (the measure is finite), whereas if $p>2$ this is the inequality $\| \cdot \|\_p \leq \|\cdot \|\_\infty^\theta \|\cdot \|\_2^{1-\theta}$ for $\theta=1-2/p>0$.


This proves that the $\| \cdot \|\_p$-closure of $A$ contains $L^\infty$, and hence it is $L^p$.


