If $\sup_n P(\vert X_n -a_n\vert > b) \leq c$, then can I find a $d \in \mathbb{R}$ such that $\sup_n P(\vert X_n \vert > d) \leq c$? Let $\{X_n\}_{n \in \mathbb{N}}$ denote a sequence of real random variables and $\{a_n\}_{n \in \mathbb{N}}$ a sequence of real numbers. Suppose $P(\vert X_n\vert>\epsilon) \to 0$ for $n \to \infty$ for all $\epsilon>0.$
If $\sup_n P(\vert X_n -a_n\vert > b) \leq c$, then can I find a $d \in \mathbb{R}$ such that $\sup_n 
 P(\vert X_n \vert > d) \leq c$? And how to choose $d$?
So I know $X_n$ converges to $0$ in probabillity. Can I use that it is a deterministic limit to say something about convergence almost surely? Or can I use
$$
\sup_n P(\vert X_n \vert  > b+ \vert a_n\vert) = \sup_n P(\vert X_n \vert - \vert a_n\vert > b) \leq \sup_n P(\vert X_n -a_n\vert > b) \leq c
$$
somehow?
 A: Actually, you do not need the assumption $\sup_n P(\vert X_n -a_n\vert > b) \leq c$ because convergence in probability is enough. Indeed, using $\varepsilon=1$ in the definition of convergence in probability, we can find $n_0$ such that for each $n\geqslant n_0$, $\mathbb P\left(\lvert X_n\rvert>1\right)\leqslant c$. For $1\leqslant n\leqslant n_0$, take $d_n>0$ such that $\mathbb P\left(\lvert X_n\rvert>d_n\right)\leqslant c$ [this follows from the fact that for any random variable $X$, $\mathbb P(\lvert X\rvert>R)\to 0$ as $R$ goes to infinity]. Finally, take $d=\max\{1,d_1,\dots,d_{n_0-1}\}$.
A: Okay I think Davide answers my question but I do have a follow up question. What if instead the setup was:
Let $\{X_{nk} \mid n \in \mathbb{N}, k \leq n \}$ denote a traingular array of real random variables and $\{a_n\}_{n \in \mathbb{N}}$ a sequence of real numbers. Suppose $\max_k P(\vert X_{kn}\vert>\epsilon) \to 0$ for $n \to \infty$ for all $\epsilon>0.$
If $\sup_n \sum_{k=1}^n P(\vert X_{kn} -a_n\vert > b) \leq c$, then can I find a $d \in \mathbb{R}$ such that $\sup_n 
 \sum_{k=1}^n  P(\vert X_{nk} \vert > d) \leq c$? And how to choose $d$?
As far as I can tell you need a different solution since this with $d=\max\{1,d_1,\dots,d_{n_0 -1}\}$ would yield the bound
$$
\sum_{k=1}^n  P(\vert X_{nk} \vert > d) \leq nc
$$
which doesn't give anything useful when applying $\sup_n$ on both sides.
