Showing that $\bigvee\limits_{i\in I}\mathcal A_i=\bigcup\limits_{I_0\subset I,\text{ countable}}\bigvee\limits_{i\in I_0}\mathcal A_i$.

Question

Let $\{\mathcal A_i:i\in I\}$ be an arbitrary collection of $\sigma$-fields in a measurable space $(\Omega,\mathcal A)$.We define, $\bigvee\limits_{i\in I}\mathcal A_i=\sigma(\bigcup\limits_{i\in I} \mathcal A_i).$Now with this definition I am asked to solve the following result:

Show that $A\in \bigvee\limits_{i\in I}\mathcal A_i$ if and only if $A\in \bigvee\limits_{i\in I_0}\mathcal A_i$ for some countable subset $I_0\subset I$.

So basically we are trying to prove that $\bigvee\limits_{i\in I}\mathcal A_i=\bigcup\limits_{I_0\subset I \text{ countable}}\bigvee\limits_{i\in I_0} \mathcal A_i$.Now one side is obvious which is the RHS $\subset$ LHS part.But I am having a problem to do the other way.Some help from MSE members will be highly appreciated.

Answer 1

Hint: It is fairly easy to verify that RHS is a $\sigma-$ algebra. Since it contains each $\mathcal A_i$, it contains LHS.