# $X_n \to X$ a.s. iff $\mathbb{P}(|X_n-X|>\epsilon_n \, \, \text{i.o.})=0$ for some $\epsilon_n \to 0$

I know that $$X_n \rightarrow X$$ a.s. $$\leftrightarrow$$ $$P(|X_n-X|>\epsilon\ \text{i.o.})=0$$ for all $$\epsilon>0$$.

My question is : If we show $$P(|X_n-X|>\epsilon_n\ \text{i.o.})=0$$ such that $$\epsilon_n$$ goes to zero, can we conclude that $$X_n$$ converges to $$X$$ almost surely? How about the converse?

I've seen that by $$P(|X_n-X|>\frac{1}{n}\ \text{i.o.})=0$$ people conclude that $$X_n$$ converges to $$X$$ a.s., but I'm not sure about a general $$\epsilon_n$$.

If $$\mathbb{P}(|X_n-X|> \epsilon_n \, \, \text{i.o.}) = 0, \tag{1}$$ then $$X_n \to X$$ almost surely. This follows simply from the fact that $$(1)$$ means that for almost every $$\omega \in \Omega$$ we can find $$N \in \mathbb{N}$$ such that $$\omega \notin \{|X_n-X|>\epsilon_n\} \quad \text{for all n \geq N,}$$ i.e. $$|X_n(\omega)-X(\omega)| \leq \epsilon_n \quad \text{for all n \geq N.}$$ As $$\epsilon_n \to 0$$ as $$n \to \infty$$ we get $$\lim_{n \to \infty} |X_n(\omega)-X(\omega)|=0$$.

The converse does in general not hold true, i.e. $$X_n \to X$$ almost surely does in general not imply $$(1)$$ for an arbitrary sequence $$\epsilon_n \to 0$$. Consider for instance $$X_n := 2/n$$ and $$\epsilon_n = 1/n$$, then $$X_n \to X=0$$ almost surely but $$\mathbb{P}(|X_n-X|> \epsilon_n \, \, \text{i.o.}) = 1.$$ However, if $$X_n \to X$$ a.s. then we can always find a sequence $$\epsilon_n \to 0$$ such that $$(1)$$ holds. Indeed: As $$X_n \to X$$ almost surely we have

$$\mathbb{P} \left( \sup_{n \geq N} |X_n-X| \geq \epsilon \right) \xrightarrow[]{N \to \infty} 0$$

for any $$\epsilon>0$$. Choosing $$\epsilon = 1/k$$ for fixed $$k \in \mathbb{N}$$ this means that there exists $$N_k \in \mathbb{N}$$ such that

$$\mathbb{P} \left( \sup_{n \geq N_k} |X_n-X| \geq \frac{1}{k} \right) \leq \frac{1}{k^2}.$$

Without loss of generality, we may assume that $$N_1 < N_2 < \ldots$$. Applying the Borel-Cantelli lemma we find that

$$\mathbb{P} \left( \sup_{n \geq N_k} |X_n-X| \geq \frac{1}{k} \, \, \text{for infinitely many k} \right)=0.$$

If we define

$$\epsilon_n := \frac{1}{k} \qquad \text{for} \, \, N_k \leq n < N_{k+1}, k \in \mathbb{N}$$

then this shows

$$\mathbb{P}(|X_n-X| \geq \epsilon_n \, \, \text{i.o.})=0.$$

• You gave a counter example for a specific sequence of $\epsilon_n$. Is there a sequence of $\epsilon_n$ such that $P(|X_n-X|>\epsilon_n i.o.)=0$? Oct 5, 2018 at 13:48
• @S_Alex For the particular example which I gave you can answer this question yourself, can't you?
– saz
Oct 5, 2018 at 15:01
• I want to construct a sequence $1,...,1,\frac{1}{2},....,\frac{1}{2},\frac{1}{3},...,\frac{1}{3},....$. But I'm not sure exactly when to switch from $1$ to $\frac{1}{2}$, from $\frac{1}{2}$ to $\frac{1}{3}$, etc Oct 5, 2018 at 15:32
• @S_Alex Not sure what you are talking about... you could simply put e.g. $\epsilon_n := 3/n$
– saz
Oct 5, 2018 at 18:15
• I'm talking about the general case. When $X_n$ is a general random sequence and it converges to $X$ almost surely. Then we can find a sequence $\epsilon_n$ such that $P(|X_n-X|>\epsilon_n \ i.o.)=0$ and $\epsilon_n \rightarrow 0$ as $n \rightarrow \infty$. Oct 5, 2018 at 18:19