Let $\mu_1,\mu_2 \in \mathcal{M}(\Omega)$ be such that $\mu_2 \ge \mu_1 \ge 0.$ It is clear that $$\|\mu_2\| \ge\|\mu_1\|,$$ where $$\|\mu\| = \sup \left\{ \int_{\Omega} \varphi d\mu: \varphi \in C_0(\overline{\Omega}) \text{ and }|\varphi| \le 1 \right\}.$$ However, I wonder is the converse true: If $\|\mu_2\| = \| \mu_1\|$ and $0 \le \mu_1 \le \mu_2$, it is true that $\mu_1 = \mu_2$?
Thanks for your helps.
The statement is false without an assumption of finiteness; for example, $dx$ and $\frac 1 2 dx$ on $\mathbb{R}$.
By the Radon-Nikodym theorem, there is a function $f : \Omega \to [0, 1]$ with $$d\mu_1 = f d\mu_2$$
and the mass of $\mu_1$ is given by $\|f\|_{L^1(\mu_2)}$. If there were a set $E$ for which $\mu_1(E) < \mu_2(E)$ then we would have $f < 1$ on a set $E$ of positive $\mu_2$-measure. But then
$$\|\mu_1\| = \int_E f \, d\mu_2 + \int_{\Omega \setminus E} f \, d\mu_2 < \mu_2(E) + \mu_2(\Omega \setminus E) = \|\mu_2\|$$ gives a contradiction. Hence $f \equiv 1$ almost everywhere with respect to $\mu_2$, and hence with respect to $\mu_1$. It follows that $\mu_2 = \mu_1$.