A Stochastic Limiting Inequality Proof Let $(X_p)_{p\ge 0}$ be a sequence of non-negative random variables with finite mean for each $p\ge 0$. Then 
$$\liminf_{p\to\infty} X_p^{\frac{1}{p}}\le \liminf_{p\to\infty}E(X_p)^{\frac{1}{p}}$$
almost surely.
This is a lemma in a paper I am reading, specifically Lemma 3.2. of http://www.statslab.cam.ac.uk/~mike/papers/parallel-shifts.pdf. I am a little miffed at the motivation of the proof. I would like to see what alternative proofs there are and hopefully understand what could inspire a proof for this kind of propositions.
 A: Okay, I had a look into the paper and the first thing to notice is that it is easy to proof that $E(X)<=x$.
So let us first do that and see what happens:
Define (just like in the paper) : $X=\lim \inf_{p \rightarrow \infty} X_p^{\frac{1}{p}}, x= \lim \inf_{p \rightarrow \infty} E(X_p)^{\frac{1}{p}}$ and notice that Fatou's Lemma states that $E(X) \leq E(\lim \inf_{p \rightarrow \infty} X_p^{\frac{1}{p}} \leq \lim \inf_{p \rightarrow \infty} E(X_p)^{\frac{1}{p}} = x (=E(x))$.
So that means that $E(X-x) \leq 0$.
But that is not enough for convergence P-almost surely.
So now consider when something converges P-almost surely. 
We want to proof that : $P(X \leq x)=1$. One of the ways to do that is by considering the space where $X >x$. So we actually want to show that the measure of this space becomes $0$.
That is a difficult statement, so let us consider the expectation of our variable $X-x$. We are then sure that $E((X-x) \mathbb{1}_{X>x}) \geq 0$, cause we only count positive values then. 
The important thing is to write down $\lim \inf_{p \rightarrow \infty} E(X_p^{\frac{1}{p}} \mathbb{1}_{X>x})$ and notice that
 they are not independent, so we have to split them and we can do that by using Holder's inequality, which says that 
$\lim \inf_{p \rightarrow \infty} E(X_p^{\frac{1}{p}} \mathbb{1}_{X>x}) \leq x E(\mathbb{1}_{X>x})^{1-\frac{1}{p}}$.
Now the arguments are split and the rest is solvable.
A: I am not sure this qualifies for an alternative proof anyway, define 
$$A_j:=\left \{ \liminf_p X_p^{ 1/p}\geqslant \liminf_p\left(\mathbb E[X_p]\right)^{1/p} +\frac 1j\right\}.$$
We have to prove that for each $j$, $\mathbb P(A_j)=0$. If not, then by Markov's inequality and Fatou's lemma, 
$$\liminf_p\left(\mathbb E[X_p]\right)^{1/p} +\frac 1j\leqslant 
\frac 1{\mathbb P(A_j)}\int_{A_j}\liminf_pX_p^{ 1/p}\mathrm d\mathbb P 
\leqslant \liminf_p\int_{A_j}X_p^{ 1/p}\frac {\mathrm d\mathbb P}{\mathbb P(A_j)} .$$
Since the map $t\mapsto t^{1/p} $ is concave and $\mathrm d\mathbb P/\mathbb P(A_j)$ is a probability measure, we obtain 
$$\liminf_p\left(\mathbb E[X_p]\right)^{1/p} +\frac 1j\leqslant \liminf_p\left(\int_{A_j}X_p\frac {\mathrm d\mathbb P}{\mathbb P(A_j)}\right)^{ 1/p}\leqslant \liminf_p\left(\int_{\Omega}X_p\frac {\mathrm d\mathbb P}{\mathbb P(A_j)}\right)^{ 1/p}=\frac{\liminf\limits_p\left(\mathbb E[X_p]\right)^{1/p}}{\liminf\limits_p\mathbb P(A_j)^{1/p}}=\liminf_p\left(\mathbb E[X_p]\right)^{1/p},
$$
which is a contradiction.
A: This looks a lot like Fatou's lemma: http://en.wikipedia.org/wiki/Fatou%27s_lemma
Maybe by defining f_p =X_p^(1/p) and letting \mu be the Lebesgue measure, you can already obtain the result.
A: Normally, we define $(\Omega, \sigma-\text{algebra}, P)$ as a probability space and here we define $(X_p)_{p \geq 0}$ on that space with $(X_p)\geq 0$. We write then $E(X_p) = \displaystyle\int_\Omega X_p dP$. Now define $Y_p:=X_p^{1/p}$. 
So the usual statement of Fatou's Lemma becomes then 
$E(\liminf Y_p)\le\liminf E(Y_p)$.
Now note that the difference here is that the question require you to say something about the convergence in mean. So define again $Z_p=\liminf Y_p$ and the statement then becomes $Z_p\le E(Z_p)$ in $P$-a.s.
Now note that you have a non-negative r.v. with finite mean, thereby this can be shown (the space where the distance is more than $\epsilon$ has measure $0$).
