Probabilistic inequality with two random variables The problem:

Let $\xi,\eta$ be two independent integrable r.v. such that $\mathsf P\{\xi> 0\} = 1$ and $\mathsf P\{\eta\geq 0\} = 1$ and 
  $$
a = \mathsf E[\xi-\eta]>0.
$$
  Check if 
  $$
\lim\limits_{k\to\infty}\mathsf P\{(\xi-\eta)+k\xi\geq y\} = 1
$$
  for any fixed $y$.

I can neither prove that the limit is $1$ nor find a counterexample, so any help is appreciated.
What I've tried so far: since $a>0$ then $p = \mathsf P\{\xi-\eta\geq a\}>0$. Then for a fixed $y$ we have:
$$
\mathsf P\{(\xi-\eta)+k\xi\geq y\} \geq p\cdot \mathsf P\{a+k\xi\geq y|\xi-\eta\geq a\}
$$
but even if $\mathsf P\{a+k\xi\geq y|\xi-\eta\geq a\}\to1$ with $k\to\infty$ it woulnd't be sufficient for the original problem.
 A: I don't think independence is necessary. Since $E[\xi-\eta]>0$, $P(\eta=\infty)=0$. So, for almost every $\omega$, by archimedean property we can find an integer $k$ such that $(k+1)\xi(\omega)\ge \eta(\omega)+y$. 
Hence $$P\left(\bigcup_{k=1}^{\infty}\{(\xi-\eta)+k\xi\ge y\}\right)=1$$
Since the events in the union are increasing in $k$, we have $$\lim_{k\to \infty}P\{(\xi-\eta)+k\xi\ge y\}=1$$
A: Since $\mathsf{P}\{\xi> 0\} = 1$, we have
$$
\lim_{k\to\infty}\mathsf{P}\left\{\xi>\frac{y+\frac{2}{\epsilon}\mathsf{E}|\xi-\eta|}{k}\right\}=1\tag{1}
$$
for any $\epsilon>0$. Furthermore, Markov's Inequality implies that
$$
\mathsf{P}\left\{\xi-\eta<-\frac{2}{\epsilon}\mathsf{E}|\xi-\eta|\right\}\le\mathsf{P}\left\{|\xi-\eta|>\frac{2}{\epsilon}\mathsf{E}|\xi-\eta|\right\}\le\frac{\epsilon}{2}\tag{2}
$$
Note that
$$
\begin{align}
\mathsf{P}(x+y\ge A+B)
&\ge\mathsf{P}(x\ge A\wedge y\ge B)\\
&=\mathsf{P}(x\ge A)-\mathsf{P}(x\ge A\wedge y<B)\\
&\ge\mathsf{P}(x\ge A)-\mathsf{P}(y<B)\tag{3}
\end{align}
$$
Therefore, given $\epsilon>0$, $(1)$ insures that there is a $k$ so that
$$
\mathsf{P}\left\{\xi>\frac{y+\frac{2}{\epsilon}\mathsf{E}|\xi-\eta|}{k}\right\}\ge1-\frac{\epsilon}{2}\tag{4}
$$
then $(2)$, $(3)$, and $(4)$ yield
$$
\begin{align}
\mathsf{P}\{(\xi-\eta)+k\xi>y\}
&\ge\mathsf{P}\{k\xi>y+\frac{2}{\epsilon}\mathsf{E}|\xi-\eta|\}-\mathsf{P}\{\xi-\eta<-\frac{2}{\epsilon}\mathsf{E}|\xi-\eta|\}\\
&\ge1-\epsilon\tag{5}
\end{align}
$$
Therefore, $(5)$ tells us that
$$
\lim_{k\to\infty}\mathsf{P}\{(\xi-\eta)+k\xi>y\}=1
$$
A: Inspired by the answer by Ashok, I also derived bounds for the convergence which I need as well. Just for the case I put it here as an answer. 
Let us take $F_\xi$ to be a c.d.f. of $\xi$. Consider
$$
1-\mathsf P\{(\xi-\eta)+k\xi\geq y\} = \mathsf P\{\eta > (k+1)\xi -y\}.
$$
We have then:
$$
\mathsf P\{\eta > (k+1)\xi -y\} = \int\limits_0^\infty \mathsf P\{\eta > (k+1)x -y\}\,dF_\xi(x)$$
$$ 
= \int\limits_0^{\frac{y+\sqrt{k+1}}{k+1}} \mathsf P\{\eta > (k+1)x -y\}\,dF_\xi(x)+\int\limits_{\frac{y+\sqrt {k+1}}{k+1}}^\infty \mathsf P\{\eta > (k+1)x -y\}\,dF_\xi(x).
$$
The first term is $\displaystyle{F_\xi\left(\frac{y+\sqrt{k+1}}{k+1}\right)}$ and the second we bound by Markov inequality:
$$
\int\limits_{\frac{y+\sqrt {k+1}}{k+1}}^\infty \mathsf P\{\eta > (k+1)x -y\}\,dF_\xi(x)\leq \mathsf E\eta\int\limits_{\frac{y+\sqrt {k+1}}{k+1}}^\infty \frac{dF_\xi(x)}{(k+1)x-y}.
$$
Since for the denominator we have: $(k+1)x-y\geq \sqrt{k+1}$ for all $x$ in the domain of integration,
$$
\int\limits_{\frac{y+\sqrt {k+1}}{k+1}}^\infty \mathsf P\{\eta > (k+1)x -y\}\,dF_\xi(x)\leq \frac{\mathsf E\eta}{\sqrt{k+1}}
$$
so
$$
\mathsf P\{\eta > (k+1)\xi -y\}\leq F_\xi\left(\frac{y+\sqrt{k+1}}{k+1}\right)+\frac{\mathsf E\eta}{\sqrt{k+1}}
$$
which in particular means that $\mathsf P\{(\xi-\eta)+k\xi\geq y\}\to 1$ with $k\to\infty$.
