**Abstract:** The classical Lebesgue density theorem states that for any Lebesgue measurable $E \subset [0,1]$ and $\mathcal{L}$-almost every $x \in E$, $\lim_{r \to 0} \frac{ \mathcal{L}(E \cap B_r(x))}{\mathcal{L}(B_r(x))} = 1$. A typical way to prove this uses a maximal inequality, which in turn uses a weak Vitali covering lemma and that fact that $\mathcal{L}$ is doubling, meaning $\sup_{x \in [0,1]} \sup_{r > 0} \frac{\mathcal{L}(B_{2r}(x))}{\mathcal{L}(B_r(x))} < \infty$. The statement of the density theorem has a clear generalization to any metric measure space and can be proven true in any doubling space by proving a stronger Vitali covering lemma. In this talk, we'll work only with measure spaces and won't consider any metric or topological structure. The sets $\{B_r(x)\}_{r >0}$ willbe generalized to nets of measurable sets $\{B_\alpha(x)\}_{\alpha \in A}$ that "converge" to $x$. We then show that the stronger Vitali covering lemma is actually equivalent to the density theorem in this setting. An application will include an alternate proof of the almost sure convergence of uniformly bounded martingales.