I was solving this exercise in a book, "A First Look at Rigorous Probability Theory", by Jeffrey Rosenthal.
Exercise 2.7.3. Suppose $\mathcal{F}$ is a collection of subsets of $\Omega$, such that $\Omega \in \mathcal{F}$. b) Assume $\mathcal{F}$ is a semialgebra. Prove that $\mathcal{F}$ is an algebra.
However, it seems to me this exercise is wrong. A counterexample suffices to show it. Suppose a set $\mathcal{J}$ of all intervals in $[0,1]$. Then, it is easy to show that $\mathcal{J}$ is a semi-algebra. Suppose two different intervals $A = [0,\frac{1}{3}]$ and $B = [\frac{2}{3},1]$ are present in $\mathcal{J}$. $A \cup B$ does not belong to $\mathcal{J}$ because it is not an interval. $\mathcal{J}$ is not an algebra because it is not closed under finite unions.
Am I missing something?