Why does $P(\limsup_n |X_n-X|>0)\le P(|X_n-X|>\frac 1 k \text{ i.o. for some } k\in \mathbb{N})$ hold?

Question

Let $X, X_n$ be $\mathbb{R}^d$-valued random variables. The solution for an exercise uses the following step without further comment: $$P(\limsup_n |X_n-X|>0)\le P(|X_n-X|>\frac 1 k \text{ i.o. for some } k\in \mathbb{N})$$ As far as I know, I should however be able to rewrite $\limsup$ on the left as follows, which seems a bit of a jump: $$P(\limsup_n |X_n-X|>0) = P(|X_n-X|>0 \text{ i.o.}) \le P(|X_n-X|>\frac 1 k \text{ i.o. for some } k\in \mathbb{N})$$ Can somebody please explain why this step is allowed?

Answer 1

$\{\limsup_n |X_n - X| > 0\} \subseteq \{|X_n - X| > 0 \text{ i.o.}\}$, but equality does not hold.

• Showing the inclusion: if the limsup is positive, then there is a subsequence $(|X_{n_k} - X|)_k$ that converges to a positive number. Thus, $|X_{n_k} - X| > 0$ i.o. and thus $|X_n - X| > 0$ i.o.
• Showing the reverse inclusion need not hold: If $|X_n - X| = \frac{1}{n}$, then $|X_n - X| > 0$ always, but the lim sup is zero. Note that this is also a counterexample to the false claim $\{|X_n - X| > 0 \text{ i.o.}\} \subseteq \{|X_n - X| > 1/k \text{ i.o. for some $k$}\}$

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To show $\{\limsup_n |X_n - X| > 0\} \subseteq \{|X_n - X| > 1/k \text{ i.o. for some $k$}\}$, modify the argument in first bullet point slightly.

Answer 2

As stated, this is wrong unless you add some more conditions. Consider $X \not=0$ a.s. and consider the random variables $X_1 = 0$ and $X_n = X$ for $n\geq 2$. Then $\mathbb{P}(\limsup_n |X-X_n| > 0) = \mathbb{P}(|X-X_1|>0)=\mathbb{P}(|X|>0)=1$. But if $n\geq 2$, then $|X-X_n| = 0$ a.s. and for any $k$ we have that $\mathbb{P}(|X-X_n|>\frac{1}{k}) = 0$. So your statement is wrong because this fact holds for all $k$.

Answer 3

$P(\limsup_{n\to\infty}|X_n−X|>0) = P(\inf_{n\to\infty} \sup_{m\ge n} |X_m-X|>0) \le P(\sup_{n\to\infty} |X_n -X|>0) = P(\sup_{n\to\infty}|X_n−X|> \varepsilon ,\varepsilon> 0)\le P(\sup_{n\to\infty}|X_n−X|> 1/k ,1/k<\varepsilon) \le P(|X_n-X| > 1/k,i.o.,\text{ for some very large }k).$