Let $(A,\mathcal{F},\mu)$ be a measure space and $f$ a measurable real
function. Define $\text{ess} \sup (f)=\inf\{c\in \mathbb{R}:\mu(\mid f\mid>c)=0\}$ and $\text{ess} \inf (f)=\sup\{c\in \mathbb{R}:\mu(\mid f\mid<c)=0\}$.
Show that $\text{ess} \sup (f+c)=\text{ess} \sup (f)+c, \forall c \in \mathbb{R}$.
So far I know that $\text{ess} \sup (f+c)\leq\text{ess} \sup (f)+c$.
Best Answer
False! If $f=-c$ ($c>0$) then $esssup(f+c)=0$ and $esssup(f)+c=2c$. If you remove absolute value signs in the definition of esssup then the result is true.