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Let $\rho_1$, $\rho_2$ be two measures(not necessarily nonnegative) on $(\Omega,\mathcal{F})$, where $\Omega$ is a set, and $\mathcal{F}$ is a $\sigma$-field in $\Omega$. Let $\mathcal{F}_0$ be a field in $\Omega$ and assume that $\mathcal{F}$ is the $\sigma$-field generated by $\mathcal{F}_0$. Suppose that $\rho_1$, $\rho_2$ are $\sigma$-finite over $\mathcal{F}_0$. (Namely, we can find a countable partition $\cup\Omega_j$ of $\Omega$, where each $\Omega_j\in\mathcal{F}_0$ and $\rho_1(\Omega_j)<\infty$, $\rho_2(\Omega_j)<\infty$.)

If $\rho_1(A)\le \rho_2(A)$ for any $A\in \mathcal{F}_0$, then do we have $\rho_1(A)\le \rho_2(A)$ for any $A\in \mathcal{F}$?

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    $\begingroup$ Consider the set of all $A$ that satisfy $\rho_1(A)\le\rho_2(A)$ and show that it is a $\sigma$-algebra. $\endgroup$ – user1688 Nov 4 '17 at 5:00
  • $\begingroup$ @Corbennick: In general, this set is not a $\sigma$-algebra. However, I think it is a monotone class (at least when you restrict to the case of finite measures, and then the $\sigma$-finite case follows easily). $\endgroup$ – Nate Eldredge Nov 4 '17 at 14:33
  • $\begingroup$ I took away the probability-measures tag since you said you want $\sigma$-finite signed measures. $\endgroup$ – Nate Eldredge Nov 4 '17 at 14:39
  • $\begingroup$ Also, I am guessing the $\sigma$-finite assumption should say that $|\rho_i|(\Omega_j) < \infty$, etc? $\endgroup$ – Nate Eldredge Nov 4 '17 at 14:40
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This is tailor-made for the monotone class theorem.

Suppose first that $\rho_1, \rho_2$ are finite signed measures. Then the collection $\mathcal{M} = \{A \in \mathcal{F} : \rho_1(A) \le \rho_2(A)\}$ is easily seen to be a monotone class (because countably additive finite signed measures are continuous from above and from below). By the monotone class theorem, since $\mathcal{F}_0 \subset \mathcal{M}$ by assumption, we conclude $\mathcal{F} = \sigma(\mathcal{F}_0) \subset \mathcal{M}$, which is the desired statement.

In the $\sigma$-finite case, consider the finite measures $\rho_i^j(A) = \rho_i(A \cap \Omega_j)$ which are, in effect, the restriction of $\rho_i$ to $\Omega_j$. By applying the above to $\rho_i^j$, conclude that $$\rho_1(A \cap \Omega_j) = \rho_1^j(A) \le \rho_2^j(A) = \rho_2(A \cap \Omega_j)$$ for all $A \in \mathcal{F}$. Now use continuity from below once more.

(I assume $\sigma$-finite here means that the positive parts of $\rho_i$ are $\sigma$-finite and the negative parts are finite, or vice versa. If both parts are potentially non-finite then you have problems with additivity because you can get $\infty - \infty$.)

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  • $\begingroup$ Thanks! But in the $\sigma$-finite case, I don't know how to conclude that $\rho_1(A\cap\Omega_j)\le \rho_2(A\cap\Omega_j)$, though we know that $\rho_1(A)\le \rho_2(A)$ and $\rho_1(\Omega_j)\le \rho_2(\Omega_j)$. Could you add some details? Thanks! $\endgroup$ – mygreatwall Nov 4 '17 at 16:26
  • $\begingroup$ @mygreatwall: See edit. $\endgroup$ – Nate Eldredge Nov 4 '17 at 16:30

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