Consider the following theorem:
Theorem: Let $(X, \mathcal{A})$ be a measurable space, and let $\mathcal{C}$ a class of subsets of $X$. Supose that $X \in \mathcal{C}$ and $\mathcal{C}$ is closed under finite intersections, i.e $$ A \cap B \in \mathcal{C}, \quad \forall A, B \in \mathcal{C} . $$ If $\mathcal{D}$ is the smallest class that contains $\mathcal{C}$, and such that is closed under finite differences ($\forall A,B\in\mathcal{D}$, $A\backslash B\in\mathcal{D}$) and y increasing limits ($\forall\{E_n\}_{n\in\mathbb{N}}\subseteq\mathcal{D}$ such that $E_1\subseteq E_2\subseteq\ldots$, satisfy that $\bigcup_{n\in\mathbb{N}}E_n\in\mathcal{D}$), then $$ \mathcal{D}=\sigma(\mathcal{C}) $$ where $\sigma(\mathcal{C})$ is the smallest $\sigma$-algebra that contains $\mathcal{C}$.
Let $(\Omega,\mathcal{A})$ be a measurable space, $\pi,\tau$ two measures in that space such that $\pi(\Omega)=\tau(\Omega)=1$, and $\mathcal{C}$ a class of subsets of $\Omega$ such that $\pi(A)=\tau(A)$ for all $A\in\mathcal{C}$, $\mathcal{C}$ it's closed under finite intersections and $\sigma(\mathcal{C})=\mathcal{A}$. Assuming $\Omega\in\mathcal{C}$, show that $$\forall A\in\mathcal{A},\hspace{0.2cm}\pi(A)=\tau(A)$$
My attempt: Let $\mathcal{D}$ the smallest class that contains $\mathcal{C}$ such that is closed under finite differences and increasing limits, then $\sigma(\mathcal{C})\subseteq\sigma(\mathcal{D})$, and by the theorem, $\mathcal{D}=\sigma(\mathcal{C})$, this implies $\sigma(\mathcal{D})=\mathcal{D}$, in particular $\mathcal{D}$ is algebra and by Monotone Class Theorem, $\sigma(\mathcal{D})$ is the smallest monotone class that contains $\mathcal{D}$. I have already show that $M=\{A\in\sigma(\mathcal{D}):\pi(A)=\tau(A)\}$ is monotone class, but to conclude I need to show that $\mathcal{D}\subseteq M$, and I don't know how. Am I on the right track?