How to prove the following defined collection is a sigma algebra? Let $\mu$ and $\lambda$ be two measures on a $\sigma$-algbra $\mathfrak{F}$ on $\Omega$, such that $\mu (A)=\lambda(A)$ for any $A\in \mathfrak C$, where $\mathfrak C\subset\mathfrak{F}$ is a collection having property if $A$ and $B$ are in $\mathfrak C$, then so is $A\cap B$. Assume there is $A_i\in \mathfrak C$, such that $\bigcup_{i=1}^\infty A_i=\Omega$ and $\mu(A_i)<\infty$ for all $i$. Then prove $\{A\in\sigma(\mathfrak C), \mu(A)=\lambda(A)\}$ is a $\sigma$-algebra.
This is one question from my homework. At first sight, it's easy. But actually, I have tried much time. I doubt that this question might be wrong.
 A: Let
$$\mathcal{D} := \{A \in \mathcal{F}; \mu(A)=\lambda(A)\}.$$


*

*$\emptyset \in \mathcal{D}$: This follows obviously from the fact that $\mu(\emptyset) = \lambda(\emptyset)=0$.

*Let $A \in \mathcal{D}$. By assumption, we can choose a sequence of sets $(G_n)_n \subseteq \mathcal{F}$ such that $G_n \uparrow \Omega$, $\mu(G_n)=\lambda(G_n)<\infty$. (Hint: $G_n := \bigcup_{j=1}^n A_j$.) Then $$\begin{align*} \mu(G \cap A^c) &= \mu(G_n \backslash A) = \mu(G_n)-\mu(G_n \cap A) \\ &= \lambda(G_n) - \lambda(G_n \cap A) = \lambda(G_n \backslash A) \\ &= \lambda(G_n \cap A^c). \end{align*}$$ It follows from the continuity of the measures that $\mu(A^c) = \lambda(A^c)$. Hence, $A^c \in \mathcal{D}$.

*Let $(C_j)_j \subseteq \mathcal{D}$ be a sequence of disjoint sets. Using the $\sigma$-additivity of $\mu$ and $\lambda$, it is not difficult to see that $$\mu\left( \bigcup_j C_j  \right) = \lambda \left( \bigcup_j C_j \right),$$ i.e. $\bigcup_j C_j \in \mathcal{D}$.


This shows that $\mathcal{D}$ is a Dynkin-system. As $\mathcal{C}$ is $\cap$-stable, this implies in particular $\sigma(\mathcal{C}) \subseteq \mathcal{D}$, i.e. $\lambda(A) = \mu(A)$ for all $A \in \sigma(\mathcal{C})$.
