# a.s. convergence of two sequences

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 $$P(A) = P(B) = 1$$