# Bounded probability implies convergence in probability

Let $$(X_n)$$ be a sequence of random variables and $$(a_n),(b_n)$$ be two sequences of non-negative real numbers such that $$a_n\downarrow 0$$ and $$b_n\downarrow 0$$ when $$n\to\infty$$.

If for any $$t>0$$, $$P(|X_n|\geq a_n+t)\leq b_n,$$ can we conclude that $$X_n\overset{P}{\to}0$$ as $$n\to\infty$$? From the hypothesis I know that for any $$t>0$$ $$\lim_{n\to\infty}P(|X_n|\geq a_n+t)=0.$$ But I do not see why the fact that $$a_n\downarrow 0$$ implies that $$\lim_{n\to\infty}P(|X_n|\geq t)=0.$$ I was trying to use Slutsky's theorem, but I don't know anything of the convergence of $$|X_n|-a_n$$. Other thing is that for $$n$$ sufficiently large $$a_n\leq \epsilon$$ for any $$\epsilon>0$$, so $$P(|X_n|\geq a_n+t)\leq P(|X_n|\geq \epsilon+t),$$ but that doesn't help either.

Any suggestions?

Fix $$s > 0$$. We want to show that $$\mathbb{P}(|X_n| \geq s) \to 0$$ so fix $$\varepsilon > 0$$ and we aim to show that for large enough $$n$$, $$\mathbb{P}(|X_n| \geq s) < \varepsilon$$.
By the assumption applied with $$t = \frac{s}{2}$$, $$\mathbb{P}(|X_n| \geq a_n + \frac{s}{2}) \to 0$$ Combining this fact with the assumption that $$a_n \to 0$$, there is an $$N$$ such that $$n \geq N$$ implies that $$a_n < \frac{s}{2}$$ and $$\mathbb{P}(|X_n| \geq a_n + \frac{s}{2}) < \varepsilon$$. This means that if $$n \geq N$$, $$|X_n| \geq s$$ implies that $$|X_n| \geq a_n + \frac{s}{2}$$. Therefore, for $$n \geq N$$, $$\mathbb{P}(|X_n| \geq s) \leq \mathbb{P}(|X_n| \geq a_n + \frac{s}{2}) < \varepsilon$$ which gives the desired convergence.
Let $$\epsilon>0$$ be fixed and let $$t=\frac{1}{2}\epsilon$$.
Then for $$n$$ large enough we have $$a_{n}+t<\epsilon$$ so that $$P\left(\left|X_{n}\right|\geq\epsilon\right)\leq\left(\left|X_{n}\right|\geq a_{n}+t\right)\leq b_{n}$$.
This leads to $$\lim_{n\to\infty}P\left(\left|X_{n}\right|\geq\epsilon\right)=0$$.
Then the fact that this works for every $$\epsilon>0$$ allows the conclusion that $$X_{n}\stackrel{P}{\to}0$$.