This result gives a necessary and sufficient condition for $L^1$ convergence of an almost surely convergent sequence: if $X_n \to X$ a.s. and $(X_n)$ is uniformly integrable, then $X_n \to X$ in $L^1$, and conversely. Uniform integrability is the precise condition that prevents "loss of mass at infinity" — the phenomenon where probability mass escapes to the tails, so that $\mathbb{E}[X_n]$ fails to converge to $\mathbb{E}[X]$ even though $X_n \to X$ pointwise. The forward direction is the substantial implication. Uniform integrability says $\sup_n \mathbb{E}[|X_n| \mathbb{1}_{|X_n| > \alpha}] \to 0$ as $\alpha \to \infty$; this tail control, combined with Egorov's theorem (which converts a.s. convergence to near-uniform convergence on sets of large measure), yields convergence in mean. The converse direction is easier: $L^1$ convergence implies that the sequence is bounded in $L^1$ and that the tails are controlled. This theorem is the bridge between the [Almost Sure Martingale Convergence Theorem](/theorems/1157) and the [UI Martingale Convergence Theorem](/theorems/1163). The almost sure theorem gives $X_n \to X_\infty$ a.s. for any $L^1$-bounded martingale, but without uniform integrability the convergence may fail in $L^1$. The present result shows that UI is both necessary and sufficient for the upgrade, making it the central notion in the $L^1$ theory of martingales. It also appears in the proof of the [Backwards Martingale Convergence Theorem](/theorems/1165), where uniform integrability comes for free because backward martingales are always UI.