About the definition of $L^{\infty}$ norm

Question

Let $\Omega$ a limited domain in $\mathbb{R}^{n}$, the space $L^{\infty}(\Omega)=\{f: \Omega\to\mathbb{R} $ measurable $; ||f||_{L^{\infty}(\Omega)}<\infty\}$. Then if a function $f \in L^{\infty}(\Omega)$, by definition of $L^{\infty}$ norm, $f$ is bounded a.e x, we can say that $\sup\limits_{\Omega} |f(x)| <\infty$ a.e.x or the last inequality is valid for all $x \in \Omega$?

Thanks a lot.

Answer 1

Only for almost all $x$.

This follows from the definition of the "norm". One possible (equivalent) definition is that $\|f\|_\infty$ is the smallest number $M>0$ such that $|f(x)|\le M$ for almost all $x$.

Answer 2

The "functions" in $L^p$ spaces, including $L^\infty$, are actually equivalence classes, where $f$ and $g$ are equivalent if they are equal almost everywhere. So any statement you make about $f(x)$ can only be interpreted in the "almost everywhere" sense. However, for $f \in L^\infty$ you can choose a representative of the equivalence class that is bounded everywhere. Namely, start with any representative, and change the value (say to $0$) on the set of measure $0$ where this representative's absolute value $> \|f\|_\infty$.