I've encountered something called the "essential supremum" while working with $L^{p}$ spaces (in particular, for $p=\infty$).
I tried looking it up on the internet but all the definitions use concepts from Measure Theory, which I'm not familiar with. Is there a way to wrap my head around it without having to deal with measures? I really only need to understand this to show that a piecewise continuous function $f \in C(\Omega)$ is in $L^{\infty}(\Omega)$. I can't seem to understand how the $L^\infty$ space works or how it is defined.
Any help is appreciated!
You cannot avoid specifying a measure when talking about $L^{\infty}(\Omega)$ since the space depends critically on the measure you are using. Let $(\Omega, \mathcal{F}, \mu)$ be a measure space. The $L^{\infty}$ norm of a measurable function $f$ is defined as $$\lVert f \rVert_{L^{\infty}} = \inf\{M \geq 0 : |f(x)| \leq M \text{ for $\mu$-almost every }x \in \Omega\}.$$ One can show that $|f(x)| \leq \lVert f \rVert_{L^{\infty}}$ for $\mu$-almost every $x \in \Omega$, i.e. the $\inf$ is a actually minimum. $L^{\infty}(\Omega, \mu)$ is defined as the space of (equivalence classes of a.e. equal) measurable functions $f : \Omega \to \mathbb{C}$ such that $\lVert f \rVert_{L^{\infty}} < \infty$. For your purpose, I guess you are using $\mu =$ Lebesgue measure, so that is the only measure you need to care about. To show that $\lVert f \rVert_{L^{\infty}} < \infty$, you need to show that there exists $M \geq 0$ such that $|f(x)| \leq M$ for almost every $x \in \Omega$. So the procedure is very similar to the procedure for showing that $f$ is bounded, except here you are allowed exceptions on a set of measure $0$. Note that for continuous functions $f$ and $\mu$ a Borel measure (such as Lebesgue measure), if $\mu$ assigns positive measure to all open subsets of $\Omega$, then it can be shown that $\lVert f \rVert_{L^{\infty}} = \sup_{x \in \Omega}|f(x)|$. This is the case when $\Omega$ is open, or when $\Omega$ is an interval with more than one point.