TITLE: Defining a measure as a supremum
QUESTION [7 upvotes]: Let $\Sigma$ be a $\sigma$-algebra over a set $X$, and $\mu_1$ and $\mu_2$ measures in it. It can be shown that
$$ \begin{align}
\mu_\sup: &\Sigma \to [0,\infty]\\
&E \mapsto \sup_{F \in \Sigma}\, \{\mu_1 (E \cap F) + \mu_2 (E\setminus F)\}
\end{align}$$
is a measure. How do I show that it's the smallest measure $\mu$ with domain $\Sigma$ such that $\mu\geq\max\{\mu_1, \mu_2\}$?

REPLY [6 votes]: You don't need to use $\varepsilon$ to see that $\mu_{\sup}\geq\max\{\mu_1,\mu_2\}$. Since $\emptyset$ and $X$ are elements of $\Sigma$, for any $E\in\Sigma$,
$$\mu_\sup(E)=\sup_{F\in\Sigma}\,\{\mu_1(E\cap F)+\mu_2(E\setminus F)\}\geq
\left\{\begin{align}
&\mu_1(E\cap \emptyset)+\mu_2(E\setminus \emptyset)=\mu_2(E)\\
&\mu_1(E\cap X)+\mu_2(E\setminus X)=\mu_1(E)
\end{align}\right.\quad.
$$
Now let $\mu$ be a nonnegative measure defined on $\Sigma$, such that, for all $E\in\Sigma$, we have $\mu(E)\geq \max\{\mu_1(E),\mu_2(E)\}$. Let $E\in\Sigma$ and $F\in\Sigma$. Then, we have 
$$\mu_1(E\cap F)+\mu_2(E\setminus F)\leq \mu(E\cap F)+\mu(E\setminus F)=\mu(E)$$
and taking the supremum over $F\in\Sigma$ we get $\mu_{\sup}(E)\leq \mu(E)$.
What you showed with the $\varepsilon$ is that $\mu_{\sup}\leq \mu_1+\mu_2$, but the reversal inequality doesn't need to be true (to see that, take $\mu_1=\mu_2$ and $E\in\Sigma$ such that $\mu_1(E)\neq 0$).