Does convergence in probability preserve the weak inequality? 
Suppose I have two sequences of random variables $\{x_n\}$ and $\{y_n\}$ such that $x_n\leq y_n$ and $\text{plim}\;x_n=L_x$ and $\text{plim}\;y_n=L_y$, can I say $L_x\leq L_y$ (almost surely)? Does it matter if I further impose that $L_x$ and $L_y$ are nonrandom?

I tried to replicate the argument for the nonstochastic case (included below for completeness) but I have been unsuccessful.

The result in the nonstochastic setting with $l_x=\lim x_n$ and $l_y=\lim y_n$ has a short proof as follows:
Suppose $l_y<l_x$ and consider $z_n=y_n-x_n\geq 0$. Then $\lim z_n=l_y-l_x=l_z<0$ but with $\epsilon=(l_x-l_y)/2>0$, for all $n$, we have $|z_n-l_z|=z_n-l_z\geq-l_z=l_x-l_y>\epsilon$. So $l_y<l_x$ is wrong and therefore we must have $l_y\geq l_x$.
 A: We know that
$$\Bbb{P}(|x_n-L_x|>\epsilon) \xrightarrow[n \to \infty]{}0  \\
\Bbb{P}(|y_n-L_y|>\epsilon) \xrightarrow[n \to \infty]{}0   $$
This implies that for every $k$ there is a $N_k$ such that $n\geq N_k$
$$\Bbb{P}\bigg(|x_n-L_x|>\frac{1}{k}\bigg) \leq \frac{1}{2^k}  \\
\Bbb{P}\bigg(|y_n-L_y|>\frac{1}{k}\bigg) \leq \frac{1}{2^k} $$
Let $$A_{N_k} = \bigg\{\omega\mid |x_{N_k}(\omega) - L_x| > \frac{1}{k}\bigg\}\\
B_{N_k} = \bigg\{\omega\mid |y_{N_k}(\omega) - L_y| > \frac{1}{k}\bigg\}$$
Note that $\sum_k \Bbb{P}(A_{N_k}) \leq \sum_k \frac{1}{2^k} < \infty$ and $\sum_k \Bbb{P}(B_{N_k}) \leq \sum_k \frac{1}{2^k} < \infty$. By the Borel-Cantelli Lemma, $$\Bbb{P}(\omega \in A_{N_k} \text{ for infinite many }k) = 0\\\Bbb{P}(\omega \in B_{N_k} \text{ for infinite many }k) = 0$$ So there is a set $\Omega*$ such that $\Bbb{P}(\Omega^*)=1$ and $\omega \in \Omega^*$ implies that for every $k$ sufficiently large
$$|x_{N_k}(\omega)-L_x(\omega)|  \leq \frac{1}{k}\\
|y_{N_k}(\omega)-L_y(\omega)| \leq \frac{1}{k}.$$
Therefore $x_{N_k}(\omega) \to L_x(\omega)$ and $y_{N_k}(\omega) to L_y(\omega).$
Now since $x_{N_{k}}(\omega)\leq y_{N_k}(\omega)$ we conclude that $L_x(\omega) \leq L_y (\omega)$ wich means that almost surely $L_x \leq L_y$ 
