Consider the space of measurable functions bounded almost everywhere
$$
L^{\infty}(\Omega)=\left\{ x:\Omega\rightarrow\mathbb{R}\mid x\text{ is }%
\mu-\text{measurable and }\operatorname*{esssup}_{x\in\Omega}\left\vert
f\left( x\right) \right\vert <\infty\right\}
$$
endowed with the norm
$$
\left\Vert x\right\Vert _{\infty}=\operatorname*{esssup}_{x\in\Omega
}\left\vert f\left( x\right) \right\vert =\inf\left\{ \alpha>0\mid
\left\vert f\left( x\right) \right\vert \leq\alpha\text{ a.e. on }%
\Omega\right\} ,
$$
where $\Omega\subset\mathbb{R}^{n}$ is a nonempty measurable set endowed with
the Lebesgue measure, and by $x$ we mean in fact an equivalence class of
functions which coincide almost everywhere. Consider the set
$$
A=\{x\in L^{\infty}(\Omega)\mid x(t)\geq0,\text{ for almost every }t\in
\Omega\}\text{.}
$$
Do there exis any references concerning the interior of the set $A$? If it is
nonempty, does there exist a proof for that? I found at this post [https://math.stackexchange.com/questions/3548634/the-interior-of-a-set-is-in-lp], that in the case of Lebesgue space $L^p, 1\le p<\infty$, the emptyness of the set $A$ (adapted for finite $p$, of course) is clear.
Best Answer
The result you have quoted is not for $L^{\infty}$. There are lots of points in the interior of $A$: Any function $f$ with the property that for some $\epsilon >0$ we have $f \geq \epsilon$ a.e. is an interior point of $A$ because $B(f,\epsilon) \subseteq A$. [$\|f-g\|_{\infty}<\epsilon$ implies $|f-g| <\epsilon$ a.s. which implies $g = f -(f-g)>\epsilon-\epsilon >0$ a.e.]
These are the only interior points: Suppose $f$ is in the interior and $B(f, \epsilon) \subseteq A$. Then $f-\frac {\epsilon } 2 \in B(f, \epsilon) \subseteq A$ so $f-\frac {\epsilon } 2 \geq 0$ a.e. and $f \geq \frac {\epsilon } 2$ a.e.