Ive been struggling trying to really learn what the monotone class theorem is about. I got the following problem:
Let $(\Omega,F)$ a measurable space and $\mathbb{A}\subset F$ an algebra. Let $\mu$ and $\upsilon$ two $\sigma-finite$ measures over $(\Omega,F)$ such that $\mu(A)=\upsilon(A)$ for all $A\in\mathbb{A}$. Show that $\mu(A)=\upsilon(A)$ for all $A\in \sigma(A)$.
PROOF:
Let $M=\{A\in \sigma(\mathbb{A}): \mu(A)=\upsilon(A)\}$
- Let $(B_{j})_{j}\subset M$ s.t $B_{1}\subset B_{2}\subset...$
$\mu(\bigcup_{j}B_{j})=lim_{j\rightarrow \infty}\mu(B_{j})$ by continuity over $\mu$.
$lim_{j\rightarrow \infty} \mu(B_{j}))=lim_{j\rightarrow \infty} \upsilon(B_{j})$ by hypothesis.
$lim_{j\rightarrow \infty} \upsilon(B_{j})=\upsilon(\bigcup_{j}B_{j})$ by continuity of the measure.
Then $\bigcup_{j}B_{j}\in M$
- Let $(B_{j})_{j}\subset M$ s.t $B_{1}\supset B_{2}\supset...$
$\mu(\bigcap_{j}B_{j})=lim_{j\rightarrow \infty}\mu(B_{j})$ by continuity over $\mu$ and $\sigma-finite$.
$lim_{j\rightarrow \infty} \mu(B_{j}))=lim_{j\rightarrow \infty} \upsilon(B_{j})$ by hypothesis.
$lim_{j\rightarrow \infty} \upsilon(B_{j})=\upsilon(\bigcap_{j}B_{j})$ by continuity of the measure and $\sigma-finite$.
Then , $\bigcap_{j}B_{j} \in M$
Finally, M is a monotone class, therefore $M\supset M(\mathbb{A})=\sigma(\mathbb{A})$
Is this proof correct?
Please, need all the possible help you can give cause this thing of monotone class stuff is making me really hard to keep forward with my study :c!
Thanks so much in advance!