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 $A\in \mathcal{F}$?



Extension preserves the relation between two measures
This is tailor-made for the [monotone class theorem](http://www.math.ubc.ca/~feldman/m420/monotone.pdf). 


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 $\infty - \infty$.)


