Assume we have two sequences of random elements $X_{n}$ and $Y_{n}$, taking values from some Hilbert space $S$ and defined on the same probability space. Assume that $$ X_{n}\overset{a.s.}{\to} a $$ and $$ Y_{n}\overset{a.s.}{\to} b $$ One of the definitions of a.s. convergence is $X_n \overset{a.s.}{\to} a$ if for any $\varepsilon > 0$

$$ P( \liminf_{n\to\infty} A_n) =1 $$ where $A_n = \{ \omega \in \Omega: |X_{n}(\omega) - a| < \varepsilon\}$.

Next, let $B_n = \{ \omega \in \Omega: |Y_{n}(\omega) - b| < \varepsilon\}$.

Is the following correct $$ P(\liminf_{n\to\infty} (A_n\cap B_{n})) =1? $$


It's easy to see or a well known fact that for two sets $A,B$ with $P(A) = P(B) = 1$ it holds $P(A\cap B) = 1$.

Additionally from the definition of the $\liminf$ as $$\liminf_{n\rightarrow\infty} A_n:={\bigcup_{n=1}^\infty}\left({\bigcap_{m=n}^\infty}A_m\right)$$ it's an calculation using de Morgan's laws to see $$\liminf_{n\rightarrow\infty} (A_n \cap B_n) = \liminf_{n\rightarrow\infty} A_n \cap \liminf_{n\rightarrow\infty} B_n$$

So with $$A = \liminf_{n\rightarrow\infty} A_n, B = \liminf_{n\rightarrow\infty} B_n$$ we get $$P( \liminf_{n\to\infty} (A_n \cap B_n)) = P(A \cap B) =1$$ due to $P(A) = P(B) = 1$


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