Is it so that $\sigma(\cup_i \sigma(A_i)) = \sigma(\cup_i A_i)$? You are given a set $\Omega$ with some subsets $A_i \subset \Omega$, for some indices $i\in I$.
We define a generated $\sigma$-algebra $\sigma(A_i)$ to be the smallest $\sigma$-algebra that contains $A_i$.
Is it always the case that $ \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) = \sigma\left( \bigcup_{i\in I} A_i \right) $?
My intuition says yes, and it seems clear in the case where $A_i=\{a_i\},\,a_i\in\Omega$, i.e. singleton sets. Please provide a proof or a counterexample for the general case.
ADDENDUM: As was pointed out in comments and questions, I had mixed up the subsets vs sets of subsets. The question that I meant to answer is as stated below.
Consider instead collections of subsets $\mathcal A_i$. All the elements $A\in \mathcal A_i$ are subsets of $\Omega$, i.e. $A\subseteq \Omega$. The set of such collections are indexed by $i\in I$.
Is it always the case that $ \sigma\left( \bigcup_{i\in I} \sigma(\mathcal A_i) \right) = \sigma\left( \bigcup_{i\in I} \mathcal A_i \right) $?
 A: 
Given a set $\Omega$ with some subsets $A_i \subset \Omega$, for some indices $i\in I$.
We define a generated $\sigma$-algebra $\sigma(A_i)$ to be the smallest $\sigma$-algebra that contains $A_i$.
Is it always the case that $ \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) = \sigma\left( \bigcup_{i\in I} A_i \right) $?

For the question as stated, the answer is NO.
Counter-exemple:  Let $\Omega =\{1,2\}$. Let $I =\{1,2\}$ and, for each $i\in I$, let $A_i=\{i\}$.
Then, $\bigcup_{i\in I} A_i = \{1\} \cup \{2\} = \{1,2\} $. The smallest $\sigma$-algebra that contains $ \{1,2\} $ is $\{\emptyset, \{1,2\}\}$. So, we have
$$ \sigma\left( \bigcup_{i\in I} A_i \right) = \{\emptyset, \{1,2\}\} \tag{1} $$
On other hand, for each $i\in I$,
$$ \sigma(A_i)  = \{\emptyset, \{1\}, \{2\}, \{1,2\} \} $$
It is easy to see that
$$ \sigma\left( \bigcup_{i\in \Bbb N} \sigma(A_i) \right)  = 
\sigma\left(  \{\emptyset, \{1\}, \{2\}, \{1,2\} \} \right) =  \{\emptyset, \{1\}, \{2\}, \{1,2\} \} \tag{2} $$
From $(1)$ and $(2)$, $ \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) \neq \sigma\left( \bigcup_{i\in I} A_i \right) $.
Remark: If the question was

Given a set $\Omega$ with some collections   $A_i$  of subsets of $\Omega$, for some indices $i\in I$.
Is it always the case that $ \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) = \sigma\left( \bigcup_{i\in I} A_i \right) $?

Then the answer would be YES.
Proof:
For all $i \in I$,  $A_i \subseteq \sigma(A_i)$. So
$ \bigcup_{i\in I} A_i \subseteq  \bigcup_{i\in I} \sigma(A_i)$.
So, we have
$$ \bigcup_{i\in I} A_i \subseteq  \bigcup_{i\in I} \sigma(A_i) \subseteq \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right)$$
Since $\sigma\left( \bigcup_{i\in I} \sigma(A_i) \right)$ is a $\sigma$-algebra, we have
$$ \sigma\left( \bigcup_{i\in I} A_i \right)\subseteq \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) \tag{3}$$
On the other hand, for all $i \in I$,  $A_i \subseteq \bigcup_{i\in I} A_i \subseteq \sigma\left( \bigcup_{i\in I} A_i \right)$. Since $\sigma\left( \bigcup_{i\in I} A_i \right)$ is a $\sigma$-algebra, we have, for all $i \in I$,
$$  \sigma(A_i) \subseteq \sigma\left( \bigcup_{i\in I} A_i \right) $$
So
$$  \bigcup_{i\in I} \sigma(A_i)  \subseteq \sigma\left( \bigcup_{i\in I} A_i \right) $$
Since $\sigma\left( \bigcup_{i\in I} A_i \right)$ is a $\sigma$-algebra, we have,
$$ \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) \subseteq \sigma\left( \bigcup_{i\in I} A_i \right) \tag{4}$$
From $(3)$ and $(4)$, we have that
$$ \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) = \sigma\left( \bigcup_{i\in I} A_i \right) $$
A: By definition, we have
$$\sigma\left(\bigcup_{i\in I}\mathcal A_i\right)=\bigcap X,\quad\quad\sigma\left(\bigcup_{i\in I}\sigma(\mathcal A_i)\right)=\bigcap Y,$$
where
$$X=\left\{\mathcal A\mid\text{$\mathcal A$ is $\sigma$-algebra on $\Omega$ and $\bigcup_{i\in I}\mathcal A_i\subseteq \mathcal A$}\right\}$$
and
$$Y=\left\{\mathcal A\mid\text{$\mathcal A$ is $\sigma$-algebra on $\Omega$ and $\bigcup_{i\in I}\mathcal \sigma(A_i)\subseteq \mathcal A$}\right\}.$$
Hence, it suffices to prove $X=Y$. Clearly, $Y\subseteq X$, since $\mathcal A_i\subseteq\sigma(\mathcal A_i)$, so the condition in $X$ is less restrictive. Secondly, we also have $Y\subseteq X$, since $\mathcal A_i\subseteq \mathcal A$ implies $\sigma(\mathcal A_i)\subseteq\mathcal A$, as $\mathcal A$ is a $\sigma$-algebra.
