Why $\mu_1$ satisfy the following property?

Question

Let $(X,\mathcal{M}, \mu)$ be a measure space. Consider $$ \mu_1(E) = \sup\left\{\mu(F);\; F\subseteq E,\,F\in \mathcal{M}\; \text{and}\; \mu(F)<\infty\right\}, $$ for all $E \in \mathcal{M}$.

• Why $\mu_1$ is well defined?

• Why for all $E \in\mathcal{M}$ with $\mu_1(E) = \infty$, there exists $F \in\mathcal{M}$ with $F \subset E$ and $0 < \mu_1(F) < \infty$?

Answer 1

1. Your second question suggests that $\mu_1:\mathcal P(X)\to[0,\infty]$. But, on $[0,\infty]$ the supremum of any of its subsets is well defined.
2. By definition, if $\mu_1(E)>0$, there must be an $F\in\mathcal M$ with $F\subseteq E$ and $0<\mu(F)<\infty$, since the supremum of the empty set is $0$. And observe that $\mu_1(M)=\mu(M)$ for any $M\in\mathcal M$ with finite measure.