Essentially bounded function on $\mathbb{R}$ I don't really know a lot about measure (just finishing my undergrad) so I'm not really on good terms with this. So, let $L^\infty[a,b]$ denote the space of all essentially bounded functions on $[a,b]$ with the norm $\left\| f \right\|_\infty = \operatorname{ess} \sup_{x\in[a,b]} |f(x)|$. What would exactly be the difference between a bounded and an essentially bounded function (also, the difference between the supremum and essential supremum)? If I'm not actually dealing with measures, but just with real variable functions $f:[a,b]\to\mathbb{R}$, can I omit the "essentially" part and just say that $L^\infty[a,b]$ denotes the space of bounded functions, and, additionally, omit the $\text{ess}$ in the definition of the norm? I understand that this seems like a silly question, but I've tried Googleing this and found nothing useful.

An essentially bounded function $f\in L^{\infty}([a,b])$ is explicitly related to the idea of measure, so I don't think there's a way to understand this without measures. The definition can be stated that $f\in L^{\infty}([a,b])$ if there is a $g$ measurable on $[a,b], f=g$ except on a set of measure zero, and $g$ is bounded.
Two examples: $$f_1(x)=\begin{cases} x & \text{ if } x\in\mathbb{Q}\\ 0 & \text{ otherwise}\end{cases}$$Then $f_1=0$ except on a set of measure $0$, namely $\mathbb{Q}$, so $f_1$ is essentially bounded on $\mathbb{R}$, although $f_1$ is clearly not bounded on $\mathbb{R}$.
Next,
$$f_2(x)=\begin{cases} 1 & \text{ if } x\in \mathbb{Q}\\ 0 & \text{ otherwise}\end{cases}$$ Then $f_2=0$ except on a set of measure zero, so $f_2$ is also essentially bounded. Moreover, $f_2$ is also bounded, $|f_2|\leq 1$ and $\sup |f_2|=\|f_2\|=1$. However, note that $\|f_2\|_\infty\not=\|f_2\|$. So even if a function is essentially bounded and bounded, its essential bound is not necessarily equal to the bound: it can be greater or less than the bound. Thus, the essential bound and the supremum bound are fundamentally different.
