Bochner integral of a countably valued function

Form the finite simple approximants. For $N\ge 1$, put $s_N={\sum }_{n=1}^N{x}_n{\mathbf{1}}_{A_n}$. If $x_n\mathrm{\ne }0$, the finiteness of the displayed series forces $\mu (A_n)<\mathrm{\infty }$; zero levels need no finite-measure hypothesis. Thus every $s_N$ is an integrable simple function in the precise sense of [L1]. Pairwise disjointness gives $s_N(t)\to f(t)$ for every $t$: at most one summand is nonzero at any point.

Calculate the scalar approximation error. Pointwise disjointness gives $\parallel f-s_N\parallel ={\sum }_{n>N}\parallel x_n\parallel {\mathbf{1}}_{A_n}$. Applying monotone convergence in [L3] to its finite partial sums yields

Establish Bochner integrability and identify the unrestricted integral. The everywhere simple convergence in step 1.1 proves strong measurability, and step 2.1 gives integrability of the norm. Hence [L2] makes $f$ Bochner integrable. Moreover, $(s_N)$ is a defining $L^1$-simple approximation, so

Calculate every restricted integral and audit the boundary cases. [L1, L2, step 2.1, step 3.1] For measurable $E$, the functions ${\mathbf{1}}_E{s}_N$ approximate ${\mathbf{1}}_{E}f$ in $L^1$, since their error integral is at most the tail in step 2.1. The simple calculation from [L1] therefore gives ${\int }_{E}f={\sum }_n\mu (E\cap A_n)x_n$. If $E=\mathrm{\varnothing }$, if every $A_n$ is empty, or if every $x_n=0$, both sides are zero. A single nonzero level reduces to the defining simple-function formula. Infinite-measure zero levels cause no undefined product, while a nonzero level automatically has finite measure.