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 nontrivial 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 $1\leq p < \infty$?

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). 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 $\theta=12/p>0$. This proves that the $\ \cdot \_p$closure of $A$ contains $L^\infty$, and hence it is $L^p$. 

