Proof that lim sup of union equals union of lim sup My home work is: let $ A_{n},B_{n}$ be subsets of the sample space. Prove that
$$
\limsup_{n\to\infty} (A_{n}\cup B_{n}) = \limsup_{n\to\infty} A_{n}\cup\limsup_{n\to\infty} B_{n}
$$
I managed to get to this:
$$
\bigcap_{1}^{n}\bigcup_{n\geq m}^{ } A_{m}\cup \bigcap_{1}^{n}\bigcup_{n\geq m}^{ } B_{m}\ = \bigcap_{1}^{n}\bigcup_{n\geq m}^{ } A_{m}\cup B_{m}
$$
Really appreciate if anyone can help me with this
 A: Hint. Try to use the property that $x$ is in the limsup of a sequence of sets $A_n$ iff $x$ is an element of infinitely many of the sets $A_n$.
The inclusion 
$$\limsup_{n\to\infty} (A_{n}\cup B_{n}) \supseteq \limsup_{n\to\infty} A_{n}\cup\limsup_{n\to\infty} B_{n}$$
is trivial.
As regards the other one
$$\limsup_{n\to\infty} (A_{n}\cup B_{n}) \subseteq \limsup_{n\to\infty} A_{n}\cup\limsup_{n\to\infty} B_{n}$$
you can show it by contradiction.
A: We have $x \in \limsup_{n \to \infty}(A_{n} \cup B_{n})$ iff for every $n \geq 1$ there is some $N \geq n$ such that $x \in A_{N}$ or $\in B_{N}$;  but, since $x$ is by definition independent of $n$ and $N$, this is equaivalent to the statement that for every $n \geq 1$ there is some $N \geq n$ such that $x \in A_{N}$ or for every $n \geq 1$ there is some $N \geq n$ such that $x \in B_{N}$, i.e. equivalent to $x \in \limsup_{n \to \infty}A_{n} \cup \limsup_{n \to \infty}B_{n}$.
A: I think you can prove this using the distributivity laws of sets. $x \in \text{limsup} A_n \cup B_n \iff x \in \cap_n \cup_{ k \geq n} (A_k \cup B_k) \iff x \in \cap_n [ (\cup_{k \geq n} A_k) \cup (\cup_{k \geq n} B_k)] \iff x \in [\cap_n \cup_{k \geq n} A_k] \cup [\cap_n \cup_{k \geq n} B_k] \iff x \in \text{limsup} A_n \cup \text{limsup} B_n.$
