Vector measure induced by an L-one function

Obtain the vector-measure conclusions. Apply [L1] to $f$. It gives norm countable additivity of $E\mapsto {\nu }_f(E)$, absolute continuity with respect to $\mu$, and the equality $\mathrm{\mid }{\nu }_f\mathrm{\mid }(E)={\int }_E\parallel f\parallel$ for every measurable $E$. This is an equality of finite positive measures, not merely an upper estimate on $\parallel {\nu }_f(E)\parallel$.

Calculate the countably valued special case. Put $s_N={\sum }_{n\le N}{x}_n{\mathbf{1}}_{A_n}$. These are integrable simple functions and converge pointwise to $f$. Pairwise disjointness and monotone convergence in [L2] give $\int \parallel f-s_N\parallel ={\sum }_{n>N}\mu (A_n)\parallel x_n\parallel \to 0$, so [L2] makes $f$ Bochner integrable. Restricting the same approximation to $E$ and using the finite simple formula gives ${\nu }_f(E)={\sum }_n\mu (E\cap A_n)x_n$. Also $\parallel f\parallel ={\sum }_n\parallel x_n\parallel {\mathbf{1}}_{A_n}$ pointwise, so [L1] and the same scalar monotone-convergence calculation give $\mathrm{\mid }{\nu }_f\mathrm{\mid }(E)={\sum }_n\mu (E\cap A_n)\parallel x_n\parallel$.

Audit the examples at the boundaries. [L1, L2, step 1.1, step 2.1] For $E=\mathrm{\varnothing }$ both measures vanish. For $f=0$, the induced vector measure and its variation are both zero. With one nonzero level the two formulas read ${\nu }_f(E)=\mu (E\cap A)x$ and $\mathrm{\mid }{\nu }_f\mathrm{\mid }(E)=\mu (E\cap A)\parallel x\parallel$, exhibiting equality even when cancellation would make the norm of a multi-level vector sum smaller. A zero coefficient on an infinite-measure level contributes zero under the established simple-integral convention.