Separability of $L^p$ for finite $p$, and the failure at $p=\mathrm{\infty }$

Let $(X,\mathcal{A},\mu )$ be a $\sigma$-finite measure space whose $\sigma$-algebra is countably generated modulo null sets, for instance ${\mathbb{R}}^n$ with Lebesgue measure.

Separability for finite $p$. For $1\le p<\mathrm{\infty }$ the space $L^p(\mu )$ is separable: it has a countable dense subset. On ${\mathbb{R}}^n$ one may take finite rational linear combinations of indicators of boxes with rational vertices, or the continuous functions of compact support with rational data.

Failure at $p=\mathrm{\infty }$. $L^{\mathrm{\infty }}[0,1]$ is not separable. The family $\{{\mathbf{1}}_{[0,t]}:t\in [0,1]\}$ is uncountable and satisfies $\mathrm{\parallel }{\mathbf{1}}_{[0,s]}-{\mathbf{1}}_{[0,t]}{\mathrm{\parallel }}_{\mathrm{\infty }}=1$ for $s\mathrm{\ne }t$, so the open balls of radius $\frac{1}{2}$ around its members are pairwise disjoint and any dense set must meet each of them. The same argument shows ${\mathrm{\ell }}^{\mathrm{\infty }}$ is not separable.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. For the positive half: approximation of an $L^p$ function by simple functions, then of a measurable set by a finite union of boxes up to small measure, then of the coefficients by rationals, with the three errors summed by Minkowski's inequality (Holder and Minkowski inequalities in integral form ‡). For the negative half: nothing beyond the displayed computation, once $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\mathrm{\infty }}$ and the notion of a countable set (Finite, countably infinite, countable, uncountable ↗) are available.

Which page it serves. Separability is defined later in Separability: the existence of an at most countable dense subset ↗. The pair of statements above is the standard first example of a natural Banach space that is not separable, and it is also why $L^{\mathrm{\infty }}$ behaves differently from every $L^p$ with $p$ finite. It is not separable, so it has no countable dense subset to run approximation arguments on, and the duality stops at it. For $\sigma$-finite $\mu$ the space $L^{\mathrm{\infty }}(\mu )$ is itself the dual of $L^1(\mu )$; what has no counterpart is the return trip, since the dual of $L^{\mathrm{\infty }}(\mu )$ is not $L^1(\mu )$ but a space of bounded finitely additive set functions. So the identification $L^p(\mu {)}^{\ast }=L^q(\mu )$ that holds for $1<p<\mathrm{\infty }$ stops here.

Hypotheses worth stating carefully. Separability of $L^p$ is a property of the measure space and not of $p$ alone: $L^p(\mu )$ for the counting measure on an uncountable set is not separable for any $p$, since the indicators of singletons are pairwise at distance $2^{1\mathrm{/}p}$. The clean statement is the one above, with $\sigma$-finiteness and a countably generated $\sigma$-algebra as hypotheses.