I fear I may be overlooking something obvious.
In Rudin's "Real & Complex Analysis" chapter 3 on $L_p$-spaces, we get
3.7 Definition Let $X$ be a measure space with a positive measure $\mu$.
Suppose $g: X \to [0, \infty]$ is measurable. let $S$ be the set of all real $\alpha$ such that $$\mu(g^{-1}((\alpha, \infty])) = 0.$$ If $S = \emptyset$, put $\beta = \infty$. If $S \ne \emptyset$, put $\beta = \text{inf } S$.
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Call $\beta$ the essential supremum of $g$.If $f$ is a complex measurable function on $X$, we define $\left\lVert f \right\rVert _{\infty}$to be the essential supremum of $\left\lvert f \right\rvert $, and we let $L^{\infty}(\mu)$ consist of all $f$ for which $\left\lVert f \right\rVert _{\infty} < \infty$.
Question: It seems to me that with this definition, the infinity norm will be $-\infty$ whenever $X$ itself has measure $0$. Other definitions that I've seen (like https://en.wikipedia.org/wiki/L-infinity) prevent this from happening. Am I failing to interpret Rudin's definition correctly?
You should check Rudin's definition of preimage. The preimage of a function $g : X \to Y$ is usually only defined for subsets of the codomain $Y$. Hence, since by definition it only makes sense to talk about the preimage of $\left]\alpha, \infty\right]$ under $g$ when $\left]\alpha, \infty\right] \subseteq Y$, the set $S$ of all real $\alpha$ such that $\mu(g^{-1}( \left]\alpha, \infty\right])) = 0$ only contains $\alpha \ge 0$. Thus if $\mu(X) = 0$, then $S = \left]0,\infty\right]$.