# Does law of large numbers converge in $L^1$?

I've seen the law of large numbers stated mainly in two (or three) forms: $S_n/n$ converges in probability (weak law) and converges almost surely (strong law). Also, there is convergence in the $L^2$-norm for uncorrelated random variables ($L^2$ weak law).

However there is a backwards martingale proof of the strong law of large numbers (any graduate level probability theory book, for example Durrett, should have it). The important thing is that $M_{-n}:=S_n/n$ is a backward martingale, and backward martingales converge both a.e. and in the $L^1$-norm. Then, in particular, $S_n/n$ converges in the $L^1$-norm.

Does $S_n/n$ really converge in the $L^1$-norm?

• If yes, why is this never mentioned?
• If no, what is wrong with my above proof?

Yes, $\bar X_n = S_n / n$ converges to $\mu$ in $\mathscr{L}_1$. It suffices to show that $\bar X_n$ is uniformly integrable since we already know $\bar X \to \mu$ in probability. This follows very quickly from the fact that the $X_i$ are trivially uniformly integrable (just fix $\epsilon > 0$ and choose $\delta > 0$ so that if $P(A) < \delta$ then $\int_A |X_n| \ dP < \epsilon$ for all $n$ and show this $\delta$ also works for $\bar X_n$). You also need to show $\sup E|\bar X_n| < \infty$ but this follows from the triangle inequality since $X_1 \in \mathscr L_1$.
I don't know that I would say that this isn't mentioned. I know it is an exercise in Chung's book, and I would be surprised if it wasn't also in Billingsley somewhere. People don't seem to care that much about $\mathscr L_p$ convergence around these parts because a.s. convergence and convergence in $P$ seem to come up more in practical settings, but that's just the way it seems to me as a statistics student.
• The correct way to do it is to use the lemma that $E|X|<\infty$ implies that there exists a convex function $\phi$ such that $\phi(x)/x \to \infty$. And then show that $\sup_n E\phi|\bar{X}_n| < \infty$ – Daniel Li Apr 3 '19 at 16:33
• @DanielLi the $X_i$'s are uniformly integrable because they are iid; we are using more than $L_1$-boundedness. You then use that to show the average $\bar X_n$ is uniformly integrable. – guy Apr 3 '19 at 19:12