8-9

Question: If f is real-valued and measurable on E, |E|>0, define its essential infimum on E by
ess_E\,\inf f=\sup\{\alpha:|x\in E:f(x)<\alpha|=0\}.
If f\geq 0, show that ess_E\,\inf f=(ess_E\,\sup 1/f)^{-1}.
Solution: The essential infimum on E is the supremum of all \alpha such that f(x) \geq \alpha except for on a subset of measure zero. If we assume \alphais positive,
\begin{align}
ess_E\,\inf f&=\sup\{\alpha:|x\in E:f(x)<\alpha|=0\}\\
&=\sup \{\alpha^{-1}:|x\in E:f(x)<\alpha^{-1}|=0\}\\
&=\inf \{\alpha:|x\in E:f(x)<\alpha^{-1}|=0\}^{-1}\\
&=\inf \{\alpha:|x\in E:f^{-1}(x)>\alpha|=0\}^{-1}\\
&=(ess_E\,\sup 1/f)^{-1}
\end{align}
since ess_E\,\sup f=\inf \{\alpha:|x\in E:f(x)>\alpha|=0\}.