Let $(X,\mathcal{M},\mu)$ be a measure space. I am trying to prove that: $$\mu_0(E):=\sup\{\mu(F)\mid F\subset E,\ \mu(F)<\infty\}$$ is a semifinite measure. I have already proven that this is a measure, i.e., I only need to prove that the measure $\mu_0$ is semifinite. Although this is confusing because: let $X$ be any nonempty set, let $\mathcal{M}$ be any $\sigma-$algebra on $\mathcal{M}$, and define $\mu$ be an measure on the measurable space $(X,\mathcal{M})$: $$\mu(E):= \begin{cases} 0 &E =\emptyset\\ \infty &\text{otherwise} \end{cases}$$ Then $\mu(X) =\infty$ (by assumption), but there is no measurable subset of $X$ of finite nonzero measure (the only measurable subset of finite measure is the empty set).
I thought that $\mu$ is a measure (although I don't because this is seemingly producing a contradiction with the question prompt in the Folland book) because: if $\{E_j\}_{j=1}^{\infty}\subset \mathcal{M}$ are disjoint, then: $$\mu\Bigl(\bigcup_{j=1}^{\infty}E_j\Bigr):= \begin{cases} 0 &\text{ every $E_j$ is empty }\\ \infty &\text{ there exists a nonempty $E_j$} \end{cases}$$ (Note that part (b) of question 15. of Folland is answered in another post, I am asking about part (a). I finished part (b) already by the way.)
