Self Study Measure Theory: Proof Not An Algebra

$\emptyset \in \mathcal{J}$ which is true, since $(0, 1]$ is also in $\mathcal{J}$

Is it? Why would $(0,1]\in \mathcal J$ imply that $\emptyset\in\mathcal J$? You state this as if the implication is clear, but I don't see why this is the case.

$A_i = (a_i, b_i], A_j = (a_j, b_j], \text{ wlog let } 0 < a_i,a_j < b_i,b_j \leq 1$, then $A_i \cup A_j = (\min\{a_i, a_j\}, \max\{b_i, b_j\}]$

Here, again, you present a claim without proving it. In particular, you claim that two sets, $A_i \cup A_j$ and $(\min\{a_i, a_j\}, \max\{b_i, b_j\}]$, are identical, but you do not prove that they are.

In fact, they are not always identical. It is always true that $A_i \cup A_j\subseteq (\min\{a_i, a_j\}, \max\{b_i, b_j\}]$, but the converse is not always true.

Let $A_i = (a_i, b_i], A_j = (a_j, b_j], \ A_i \cap A_j = \emptyset$ then $A_j \setminus A_i = A_j \in \mathcal{J}$ and if $A_i \cap A_j \neq \emptyset$ then wlog let $0 < a_i, a_j < b_i \leq b_j \leq 1$ and $A_j \setminus A_i = (b_i, b_j]$ which is again an half open interval.

Again, you present a claim and no proof. And as above, some of the claims here are just plain wrong. If you try to prove each of the equalities you claim, you should be able to identify which are not true.