Does $|X_n|\le\Delta_n+\delta$, with $\Delta_n\overset{p}\to0$ imply $|X_n|\overset{p}\to0$?

Question

Suppose $\Delta_n\overset{p}\to0$, and for any $\delta>0$, we have $$|X_n|\le\Delta_n+\delta.$$

Can we conclude that

$$|X_n|\overset{p}\to0?$$

Here $\Delta_n\overset{p}\to0$ means that for any $\varepsilon>0$, we have $\lim_{n\to\infty}P(|\Delta_n|>\varepsilon)=0$.

Answer 1

I guess that if we want a non.trivial question, we need $\Delta_n$ to depend also on $\delta$, otherwise, as pointed out in the comments, the inequality $|X_n|\leq \Delta_n$ would hold. So what we assume is that for all fixed $\delta$, $Y_n:=\left\lvert X_n\right\rvert\leqslant \Delta_{n,\delta}+\delta$, and $\Delta_{n,\delta}\to 0$ in probability.

Then for a fixed positive $\varepsilon\gt 0$, $$ \Pr\left\{Y_n\gt \varepsilon\right\}\leqslant \Pr\left\{\Delta_{n,\varepsilon/2}+\varepsilon/2\gt \varepsilon\right\}=\Pr\left\{\Delta_{n,\varepsilon/2}\gt \varepsilon/2\right\}. $$