Show $O(1) a.s.+o(1)a.s.=O(1)a.s.$ (not easy)

Question

Let $Y_n$ and $X_n$ be two sequences of random variables.

Say that $Y_n=O(1)$ almost surely if $\exists M>0$ such that $ P(\limsup_n\{\lvert Y_n\rvert \leq M\} )=1$, and

say that $Y_n=o(1)$ a.s. if $\forall \delta>0$ we have $ P(\limsup_n\{\lvert Y_n\rvert > \delta\} )=0$.

Question If $Y_n=O(1)$a.s. and $X_n=o(1)$a.s., how would you show that $Y_n+X_n=O(1)$a.s.?

Answer 1

Note $\omega \in \limsup_{n\to\infty} \{|Y_n| \le M\}$ if and only if $|Y_n(\omega)| \le M$ for infinitely many $n$. Which actually implies $\liminf_{n\to\infty} |Y_n(\omega)| \le M$. (Although the converse doesn't quite hold, but $\liminf_{n\to\infty} |Y_n(\omega)| < M$ does imply $\omega \in \limsup_{n\to\infty} \{|Y_n| \le M\}$.) In other words, the notation is very confusing.

Next, $\omega \in \limsup_{n\to\infty} \{|X_n| > \delta\}$ if an only if $|X_n(\omega)| > \delta$ for infinitely many $n$. So $\omega \notin \limsup_{n\to\infty} \{|X_n| > \delta\}$ if and only if $|X_n(\omega)| \le \delta$ for all but finitely many $n$.

Now suppose $Y_n$ is $O(1)$ a.s. and $X_n$ is $o(1)$ a.s. Then there exists $M>0$, and for all $\delta>0$, we have that with probability 1 we have $$ |Y_n(\omega)| \le M \text{ for infinitely many $n$, and } |X_n(\omega)| \le \delta \text{ for all but finitely many $n$} .$$ But this means that with probability 1 $$ |X_n(\omega) + Y_n(\omega)| \le M + \delta \text{ for infinitely many $n$} $$ because if you remove at most a finite number of elements from an infinite set, it is still infinite.

Answer 2

The result holds even if $X_n=O_{a.s.}(1)$ because $$ \limsup_{n\to\infty}|X_n+Y_n|\le \limsup_{n\to\infty}|X_n|+\limsup_{n\to\infty}|Y_n|<\infty \quad\text{a.s.} $$