Complex $L^{\mathrm{\infty }}$ space of a locally compact group

If $f\sim f^{\mathrm{\prime }}$ and $g\sim g^{\mathrm{\prime }}$, then outside the union of their two null exceptional sets, $f+g=f^{\mathrm{\prime }}+g^{\mathrm{\prime }}$ and $\alpha f=\alpha f^{\mathrm{\prime }}$ for every $\alpha \in \mathbb{C}$. The real and imaginary parts of these sums and scalar multiples are real linear combinations of measurable functions, so [F5] shows that they remain complex measurable; therefore the displayed operations are well-defined on classes.

The essential-supremum norm is independent of the representative by [F2] and $\mathrm{\mid }f\mathrm{\mid }=\mathrm{\mid }g\mathrm{\mid }$ wherever $f=g$. Because the set of almost-everywhere bounds is upward closed, each threshold $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },\mu }+\epsilon$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty },\mu }+\epsilon$ exceeds its infimum and is itself an almost-everywhere bound. Outside the union of their null exceptional sets, [F4] gives $\mathrm{\mid }f+g\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },\mu }+\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty },\mu }+2\epsilon$. Also $\mathrm{\parallel }\alpha f{\mathrm{\parallel }}_{\mathrm{\infty },\mu }=\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },\mu }$ by [F4] and scaling the threshold set (and directly when $\alpha =0$). Letting $\epsilon \downarrow 0$ gives the triangle inequality and homogeneity, so these operations preserve the finite-essential-supremum classes.

If $\mathrm{\parallel }[f]{\mathrm{\parallel }}_{\mathrm{\infty }}=0$, the upward-closed set of almost-everywhere bounds contains every $1\mathrm{/}n>0$. Thus each measurable set $E_n=\{x:\mathrm{\mid }f(x)\mathrm{\mid }>1\mathrm{/}n\}$ is null. Their countable union is null by [F3], and outside it $\mathrm{\mid }f(x)\mathrm{\mid }\le 1\mathrm{/}n$ for every $n$, hence $f=0$ almost everywhere and $[f]=0$. The essential-supremum norm is therefore definite, so $L^{\mathrm{\infty }}(G,\mu ;\mathbb{C})$ is a complex normed vector space.