Are they different Sigma algebras generated by a semialgebra and an algegra?

Question

Are they different Sigma algebras generated by a semialgebra $A$ and an algegra $A'$ of subsets of a set $X$?

where the Algebra $A'$ is obtained by all the disjoint collection of all finite unions of sets from $A$.

If yes, any proof? (I have proved only that the sigma algebra generated by the semialgebra is a subset of the other but I cannot prove the other direction)

Thanks!!

Answer 1

If I understand you, your question is:

If $\mathcal{A}$ is a semialgebra and $\mathcal{A}'$ is the algebra of finite, disjoint unions of members of $\mathcal{A}$, then is $\sigma(\mathcal{A}) = \sigma(\mathcal{A}')$?

The answer to this question is yes. Clearly, $\sigma(\mathcal{A}) \subseteq \sigma(\mathcal{A}')$, as you have already observed.

For the other inclusion, observe that $\sigma(\mathcal{A})$ contains every finite, disjoint union of members of $\mathcal{A}$ because it is a sigma-algebra and therefore closed under finite unions. Thus, $\mathcal{A}' \subseteq \sigma(\mathcal{A})$. Since, $\sigma(\mathcal{A}')$ is the smallest sigma-algebra containing $\mathcal{A}'$, we conclude that $\sigma(\mathcal{A}') \subseteq \sigma(\mathcal{A})$.