# Two $\sigma$-finite measures on an algebra are unique; monotone class form.

I have seen the following statement in $\pi-\lambda$ system form. But not in the monotone class form. I would just like some verification that this is correct.

Problem: $(\Omega, \Sigma)$ be a measurabla space. $\mathcal{A}$ be a Boolean algebra on $\Sigma$, i.e. closed under finite intersection, union, complement, contains $\emptyset$ and $\Omega$. Then if $\mu_1, \mu_2$ are two $\sigma$-finite measures on $\Omega$ that coincide on $\mathcal{A}$, then they also coincide on $\sigma(\mathcal{A})$.

Proof: without loss of generality, suppose both measures are finite. Let $D:= \{ E \in \Sigma : \mu_1(E)=\mu_2(E)\}$. Then $D$ is closed under increasing union and decreasing intersection. Thus, $D$ is a monotone class, contains $\mathcal{A}$. Hence, $\mu_1$ and $\mu_2$ coincide on $\mathcal{A}$.

Is this correct?

• I suppose that $\mu_1, \mu_2$ coincide on $\mathcal{A}$ and you want to prove that they also coincide on $\sigma(\mathcal{A})$. If so, this is correct. – JustDroppedIn Apr 24 '18 at 15:16
• Yes, sorry, there are a number of typos. I hope it is fine after the edit. – CL. Apr 24 '18 at 16:02
• The "without loss of generality" is not clear. In saying that $\mu_1$ (for example) is $\sigma$-finite do you mean that there is a sequence $(E_n)\subset\mathcal A$ with $\mu_1(E_n)<\infty$ for each $n$ and $\cup_nE_n=\Omega$, or merely that each $E_n$ is an element of $\Sigma$? – John Dawkins Apr 28 '18 at 15:04