Finite truncations approximate null and summable sequences

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

From The sequence spaces c_0 and ell-infinity, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with absolute value in the real case and modulus $\mathrm{\mid }a+ib\mathrm{\mid }=\sqrt{a^2+b^2}$ in the complex case. A scalar sequence here is a function $x:\mathbb{N}\to \mathbb{K}$, including index zero. Let ${\mathrm{\ell }}^{\mathrm{\infty }}=\{x=(x_n{)}_{n\in \mathbb{N}}:{\sup }_{n\in \mathbb{N}}\mathrm{\mid }{x}_n\mathrm{\mid }<\mathrm{\infty }\},c_0=\{x\in {\mathrm{\ell }}^{\mathrm{\infty }}:x_n\to 0\},$ both equipped with $\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{n\in \mathbb{N}}\mathrm{\mid }{x}_n\mathrm{\mid }$. Thus $c_0$ is a specified linear subspace of the bounded-sequence space ${\mathrm{\ell }}^{\mathrm{\infty }}$. Here $x_n\to 0$ means that for every real $\epsilon >0$ there is $N\in \mathbb{N}$ such that $\mathrm{\mid }{x}_n\mathrm{\mid }<\epsilon$ for all $n\ge N$. Addition and scalar multiplication are coordinatewise. The scalar triangle inequality makes bounded sequences and null sequences linear spaces and gives the triangle inequality for the displayed supremum norm. Absolute homogeneity follows coordinatewise, and a zero supremum forces every coordinate to vanish.

From ${\mathrm{\ell }}^p$ is the $L^p$ space of counting measure, with its stated hypotheses: On $(\mathbb{N},\mathcal{P}(\mathbb{N}),\mathrm{\#})$ with counting measure, every function $f:\mathbb{N}\to \mathbb{R}$ is measurable. Writing $a_k:=f(k)$, one has $\int \mathrm{\mid }f{\mathrm{\mid }}^{p}d\mathrm{\#}={\sum }_{k=0}^{\mathrm{\infty }}\mathrm{\mid }{a}_k{\mathrm{\mid }}^p(0<p<\mathrm{\infty }),$ by the counting-measure integral dictionary, so ${\mathcal{L}}^p(\mathrm{\#})$ is exactly the usual sequence class ${\mathrm{\ell }}^p$. Also $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{k\in \mathbb{N}}\mathrm{\mid }{a}_k\mathrm{\mid },$ because a subset of $\mathbb{N}$ has counting measure zero only when it is empty. Hence the quotient by almost-everywhere equality does nothing: for counting measure on $\mathbb{N}$, equality almost everywhere means equality everywhere.