 Bochner integral

Let $B$ be a Banach space and let $f$ be a strongly measurable function on a measure space. It is Bochner integrable exactly whenStrong measurability means almost-everywhere norm approximation by simple vector-valued functions. The Bochner integral is the norm limit of their integrals when the approximation also tends to zero in the integral norm; the limit is independent of the approximating sequence.^[ref]

For $f(t)=(t,t^2)$ on $[0,1]$ with values in ${\mathbb{R}}^2$, the integral is $(1/2,1/3)$. In an infinite-dimensional Banach space, norm convergence is part of the construction and cannot be replaced by coordinatewise convergence alone. The boundfollows from the triangle inequality for simple functions and passage to the norm limit. A bounded linear operator commutes with the integral by the same argument. Strong measurability alone does not imply integrability: the scalar example $f(t)=1/t$ on $(0,1)$ fails the norm-integral condition.