Results on convergence in $r$th mean convergence In the work I am interested in (primarily statistics), most of the results I have seen about stochastic convergence deal primarily with convergence almost surely, in probability and in distribution. In contrast, I have seen much less work aiming for convergence in $r$th mean. Of course, by Markov's inequality, convergence in $r$th mean implies convergence in probability. Since convergence in $r$th mean does not imply (nor is implied by) convergence almost surely, I'm trying to understand when I'd be interested in showing convergence in $r$th mean.
My understanding of convergence almost surely is that the set of events on which the convergence does not occur has measure $0$; I can't imagine a context in which I would seek out a different (and not weaker) notion of convergence.
I'm mostly looking for examples where convergence in $r$th mean is more relevant than convergence almost surely. By "more relevant" I mean that convergence in $r$th mean is the specific result of interest.
 A: Convergence in the $r$-th mean is often more convenient to prove mathematically than almost sure convergence.
Moreover, if you use Markov's inequality, you may get a rate of convergence for convergence in probability, if you can prove a rate for the convergence in $r$-th mean.
A simple illustrative example is the weak law of large numbers:
$$
\mathbb{P}\left[\left|\frac{1}{n}\sum_{i=1}^n X_i-E[X_1]\right|>\varepsilon\right]
\leq \frac{1}{\varepsilon^2}\mathbb{E}\left[\left|\frac{1}{n}\sum_{i=1}^n X_i-E[X_1]\right|^2\right]$$
The rest of the proof aims to show that $\frac{1}{n}\sum_{i=1}^n X_i-E[X_1]$ converges in the second mean to $0$. This is done as follows:
$$
\leq \frac{1}{\varepsilon^2}\mathbb{E}\left[\frac{1}{n^2}\sum_{i,j=1}^n\left| X_i-E[X_1]\right|\left| X_j-E[X_1]\right|\right]
$$
Now we can use linearity of the expectation, which we couldn't do when working with $\mathbb{P}$ directly:
$$
=\frac{1}{\varepsilon^2}\frac{1}{n^2}\sum_{i,j=1}^n\mathbb{E}\left[\left| X_i-E[X_1]\right|\left| X_j-E[X_1]\right|\right]=\frac{1}{\varepsilon^2}\frac{1}{n}\mathbb{E}\left[\left| X_1-E[X_1]\right|^2\right]=:\frac{\sigma^2}{\varepsilon^2n}
$$
Some concluding remarks:

*

*Proving the strong law of large numbers (almost sure convergence) is much harder.

*Moreover, some proofs of the strong law use the weak law as a Lemma (e.g. the martingale argument).
So, implicitly, they use that $\frac{1}{n}\sum_{i=1}^n X_i-E[X_1]$ converges in the second mean to $0$. Indeed, convergence in $r$-th mean can be used as a stepping stone to prove almost sure convergence! (even if the implication is not direct, and of course not always true).

*The previous derivation gives you a speed of convergence, which the "almost surely" result doesn't.

