Let
- $(\Omega,\mathcal A)$ be a measurable space
- $\mathcal E\subseteq2^\Omega$ be a semiring with $\sigma(\mathcal E)=\mathcal A$
- $\mu$ be a finite measure on $\mathcal A$
- $\nu$ be a $\sigma$-finite measure on $\mathcal A$
If $$\mu(E)\le\nu(E)\;\;\;\text{for all }E\in\mathcal E\tag1\;,$$ are we able to conclude $\mu\le\nu$?
Since $\mu$ is finite, $$m:=\nu-\mu$$ is a well-defined signed measure on $\mathcal A$. By $(1)$, $\left.m\right|_\mathcal E$ is a (nonnegative) $\sigma$-finite pre-measure on $\mathcal E$ and hence we obtain the existence of a unique (nonnegative) $\sigma$-finite measure $\tilde m$ on $\sigma(\mathcal E)=\mathcal A$.
But that doesn't immediately yield $m=\tilde m$. So, how can we conclude?
Let $\Omega = \{a,b\}$, $\mathcal{A} = 2^\Omega$. Let $\mathcal{E} = \{\emptyset, \{a\}\}$ which is a semiring that generates $\mathcal{A}$. Define $\mu$ by $\mu(\{a\})=0$, $\mu(\{b\})=1$. Take $\nu = 0$ and you have a counterexample.
If you like probability measures you may also take $\nu(\{a\}) = \nu(\{b\}) = 1/2$.
If you add the assumption $\Omega \in \mathcal{E}$ then it becomes true. Let $\mathcal{F}$ be the collection of all finite disjoint unions of sets in $\mathcal{E}$. Verify that $\mathcal{F}$ is an algebra and that $\mu(E) \le \nu(E)$ for all $E \in \mathcal{F}$. Now show that the collection $\mathcal{M} = \{E \in \mathcal{A} : \mu(E) \le \nu(E)\}$ is a monotone class. We just showed $\mathcal{F} \subset \mathcal{M}$. By the monotone class theorem, we conclude $\sigma(\mathcal{F}) \subset \mathcal{M}$. But $\sigma(\mathcal{F}) \supset \sigma(\mathcal{E}) = \mathcal{A}$.