# Defining a measure as a supremum

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 $\mu\geq\max\{\mu_1, \mu_2\}$?

• I edited your LaTeX to remove the hideous instances of manual spacing with \matrix and ridiculous amounts of \qquad. Please learn to use the align environment correctly – kahen Sep 13 '11 at 12:02
• @kahen But no-one was seeing the hideousness, and now the space between the lines that define $\mu_\sup$ is a bit large. align is a good idea, though. – Luke Sep 13 '11 at 18:13

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 $\mu_1(E)\neq 0$).

• Much simpler than I thought. Thanks. I'll erase my incorrect approach from the question. – Luke Sep 13 '11 at 19:05