Bochner integral $\Bbb R^d$

Question

If I have a function $f\colon X\to\Bbb R^d$ define on some measure space $X$, does the Bochner integral exist exactly if the component integrals exist? Do they coincide? What I mean is, is it the case that $\int_Xf d\mu=(\int_Xf_1d\mu,\dots,\int_X f_d d\mu)^T ?$ Thanks

Answer 1

Yes: assume that the Lebesgue integral of each component $f^{(i)}$ exist. Then there for each $i$ a sequence $(s_n^{(i)},n\geqslant 1)$ of simple functions for which $\int_X|f^{(i)}-s_n^{(i)}|\to 0$. Calling $(e_1,\dots,e_d)$ the canonical basis of $\Bbb R^d$, and defining $s_n:=\sum_{i=1}^ds_n^{(i)}e_i$, we get a simple function such that $\int_X\lVert f-s_n\rVert d\mu\to 0$ (for the Euclidian norm, hence all the other, since we work in a finite dimensional space).

For the other direction, use the projections. This gives equivalence of existence of the Bochner integral for the vector and its coordinates.

We can check everything coincides when $f$ is simple, and extend it.