Inverse of the Third Axiom in the Definition of a Sigma Algebra $\Sigma$ is a $\sigma$-algebra if it satisfies the following three properties:

*

*$X$ is in $\Sigma$, and $X$ is considered to be the universal set in
the following context.

*$\Sigma$ is closed under complementation: If $A$ is in $\Sigma$, then
so is its complement, $A^{c}=X-A$.

*$\Sigma$ is closed under countable unions: If $A_{1}$, $A_{2}$,
$A_{3}$, ... are in $\Sigma$, then so is $A=A_{1}\cup A_{2}\cup
   A_{3}\cup\ldots$.

If the above three are our definitions.

*

*Can we show that if  $A=A_{1}\cup A_{2}\cup A_{3}\cup\ldots$ and $A$ is in $\Sigma$ then each of the $A_{i}$ are in $\Sigma$?

*Do we need any additional conditions to show this or would the three basic axioms suffice?

Notes:

*

*I found some related discussions, but am unable to see if my doubts can be cleared using the pointers from these threads. Listing them below for completeness:

Can every member of a $\sigma$-algebra be represented by a countable union of disjoint members?
Show that any element of a sigma algebra is the union of disjoint sets

*

*I was trying to understand the concept in the question asked since the corresponding property seemed to be required to show that If $E$ has $\sigma$-finite measure, then $E$ is inner regular (Rudin RCA, Definition 2.16, some related threads below):

If $E$ has $\sigma$-finite measure, then $E$ is inner regular
Rudin Real And Complex Definition 2.16
This point is clarified: If $E$ is in a sigma algebra $\Sigma$ and $E$ has $\sigma$-finite measure then it is a countable union of sets, $E_{i}$ with finite measure ($\mu\left(E_{i}\right)<\infty$). Which means the sets that make up the union of $E$ are already in $\Sigma$ since the measure $\mu$ is defined only on the members of the sigma algebra $\Sigma$. Hence the $E_{i}$ belong to the sigma algebra, $\Sigma$

*

*Though my original doubt is clarified, I am still curios about the conditions under which the reverse of the third axiom hold. It is quite possible that I am missing something very basic. Happy to delete this question, if that is the case once my doubt is clarified.

 A: In the proof you're confused by, the $\sigma$-finite condition is used to construct a specific union with this property. This does not mean that an arbitrary union needs this property.
If $(X, \Sigma)$ is a measurable space, a measure $\mu$ is called $\sigma$-finite if there is some way to write $X = \bigcup_{n=1}^\infty X_n$ where $X_n \in \Sigma$ and $\mu(X_n) < \infty$ for all $n$.
It follows from this definition that for every $A \in \Sigma$, we can write $A = \bigcup_{n=1}^\infty A_n$ where $A_n \in \Sigma$ and $\mu(A_n) < \infty$ for all $n$. Just take $A_n = A \cap X_n$. (And this is what we mean by the set $A$ being $\sigma$-finite.)
Note that the definition specifically promises us some measurable sets. We don't need the much stronger and usually false property that every way of writing $A$ as a union $\bigcup_{n=1}^\infty A_n$ will result in $A_n \in \Sigma$ for all $n$.
A: @texmex wrote "I am still curios about the conditions under which the reverse of the third axiom hold".
Let us prove the following result.

Let $X$ be a set and $\Sigma$ be a $\sigma$-algebra of subsets of $X$. Then the condition

*

*if  $A=A_{1}\cup A_{2}\cup A_{3}\cup\ldots$ and $A$ is in $\Sigma$ then each of the $A_{i}$ are in $\Sigma$
is true if and only  $\Sigma = 2^X$.

where $2^X$ means the collection of all subsets of $X$
Proof: $(\Rightarrow)$ We already have tha $\Sigma \subseteq 2^X$. So all we need to prove is that  $2^X  \subseteq \Sigma  $.
Suppose that,

*

*if  $A=A_{1}\cup A_{2}\cup A_{3}\cup\ldots$ and $A$ is in $\Sigma$ then each of the $A_{i}$ are in $\Sigma$.

Let $E$ be any element of $2^X$, that means any subset of $X$. We can take $A_1=E$, $A_2 = X$ and  $A_3=A_4=\ldots = \emptyset$. We have
$$A_{1}\cup A_{2}\cup A_{3}\cup\ldots = E\cup X = X \in \Sigma$$ so, by the condition, we have that each of the $A_{i}$ are in $\Sigma$. In particular $A_1 \in \Sigma$, which means $E \in \Sigma$. Since $E$ is an arbitrary element of $2^X$, we have that $2^X \subseteq \Sigma$. So we have $\Sigma = 2^X$.
$(\Leftarrow)$  If $\Sigma = 2^X$, then all subsets of $X$ are in $\Sigma$ and the condition

*

*if  $A=A_{1}\cup A_{2}\cup A_{3}\cup\ldots$ and $A$ is in $\Sigma$ then each of the $A_{i}$ are in $\Sigma$
becomes trivially true.
Remark: Another way to prove $(\Rightarrow)$ is by the counter-positive, which consists in showing that if $\Sigma \neq 2^X$, then there is always a counter-example to

*

*if  $A=A_{1}\cup A_{2}\cup A_{3}\cup\ldots$ and $A$ is in $\Sigma$ then each of the $A_{i}$ are in $\Sigma$
In fact, suppose $\Sigma \neq 2^X$. then there is $E \in 2^X$ such that $E \notin \Sigma$. Take $A_1=E$, $A_2 = X$ and  $A_3=A_4=\ldots = \emptyset$. We have
$$A_{1}\cup A_{2}\cup A_{3}\cup\ldots = E\cup X = X \in \Sigma$$
but $A_1=E \notin \Sigma$.
