The finitely additive integral is well-defined and isometric

Two partition representations of $s$ have the common refinement [given, L1] $(A_j\cap B_k{)}_{j,k}$. Finite additivity replaces each original summand by its sum over the refinement; because both coefficients equal $s_n$ on a nonempty cell, the two refined sums agree. Thus $I_{\nu }^0$ is well-defined and linear.

For a partition representation,

Use the fixed dyadic grid to define a finite-range quantization $q_k(x)$ [given, L2, step 2.1] with $\mathrm{\parallel }{q}_k(x)-x{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2^{-k}$ (coordinatewise in a square over $\mathbb{C}$). Step 2.1 makes $(I_{\nu }^0(q_k(x)){)}_k$ Cauchy; [L2] supplies its limit. The same estimate shows independence of any approximating finite-range sequence, linearity, uniqueness, and the bound $\mathrm{\parallel }{I}_{\nu }\mathrm{\parallel }\le \mathrm{\mid }\nu \mathrm{\mid }(\mathbb{N})$.

Given $\epsilon >0$, choose a finite partition $(A_j)$ with [given, step 3.1] ${\sum }_j\mathrm{\mid }\nu (A_j)\mathrm{\mid }>\mathrm{\mid }\nu \mathrm{\mid }(\mathbb{N})-\epsilon$. Put $c_j=1$ when $\nu (A_j)=0$ and otherwise $c_j=\overline{\nu (A_j)}\mathrm{/}\mathrm{\mid }\nu (A_j)\mathrm{\mid }$ (the same sign formula over $\mathbb{R}$). Then $\mathrm{\parallel }{\sum }_j{c}_j{\mathbf{1}}_{A_j}{\mathrm{\parallel }}_{\mathrm{\infty }}=1$ and its integral is ${\sum }_j\mathrm{\mid }\nu (A_j)\mathrm{\mid }$. Thus $\mathrm{\parallel }{I}_{\nu }\mathrm{\parallel }\ge \mathrm{\mid }\nu \mathrm{\mid }(\mathbb{N})-\epsilon$; letting $\epsilon \downarrow 0$ proves equality. Linearity in $\nu$ is immediate from the formula.