It is meant to record. See: Answer own Question

It is written as question. Have fun! :)


It is taken from the original paper: S. Bochner, Integration

It is related to: Bochner Integral: Integrability


Given a measure space $\Omega$ and a Banach space $E$.

Consider Bochner measurable functions $S_n\to F$.

Suppose Bochner integrability as $\int\|F\|\mathrm{d}\mu<\infty$.

Denote the abstract Bochner integral by $\int F\mathrm{d}\mu$.

Define the Bochner integral axiomatically by: $$\text{(L)}\quad\int(F+G)\mathrm{d}\mu=\int F\mathrm{d}\mu+\int F\mathrm{d}\mu,\,\int\lambda F\mathrm{d}\mu=\lambda\int F\mathrm{d}\mu$$ $$\text{(N)}:\quad\left\|\int F\mathrm{d}\mu\right\|\leq\int\|F\|\mathrm{d}\mu$$ $$\text{(DC)}:\quad\|F_n\|\leq h:\quad F_n\to F\implies\int F\mathrm{d}\mu\to\int F\mathrm{d}\mu\quad\left(\int h\mathrm{d}\mu<\infty\right)$$

And especially for finite measures: $$\text{(C)}:\quad\int C\mathrm{d}\mu=C\mu(\Omega)$$


This forces the integral for simple functions to become: $$S=\sum_{k=1}^KS_k\chi(A_k):\quad\int S\mathrm{d}\mu=\sum_{k=1}^KS_k\mu(A_k)$$

Now, for Bochner measurable functions Bochner integrability should determine the integral, too: $$S_n\to F\implies\int S_n\mathrm{d}\mu\to\int F\mathrm{d}\mu$$

But how to apply dominated convergence therefore: $$\|F-S_n\|\to0\implies\|S_n\|\leq\|F\|$$ (I got it now; in any way, you are warmly encouraged to do give an answer, too!!)


Introduce the restricted integral through $\int_A F\mathrm{d}\mu:=\int\chi_AF\mathrm{d}\mu$.

There is another axiom which seems not independent of the others: $$\text{(S)}:\quad\int_A F\mathrm{d}\mu=\sum_{k\in\mathbb{N}}\int_{A_k}F\mathrm{d}\mu\quad\left(A=\bigsqcup_{k\in\mathbb{N}}A_k\right)$$ (Is it really independent?)

  • $\begingroup$ Is the reference really Bochner's original paper? $\endgroup$ – C-Star-W-Star Nov 28 '14 at 23:15
  • $\begingroup$ Isn't the axiom on disjoint sums redundant? (S) $\endgroup$ – C-Star-W-Star Nov 28 '14 at 23:18
  • $\begingroup$ Dude, you don't have to ask yourself questions in the comments also. $\endgroup$ – Michael Nov 29 '14 at 0:05
  • $\begingroup$ @Michael: Haha ^^ no these are really questions. $\endgroup$ – C-Star-W-Star Nov 29 '14 at 1:12

One can choose an appropriate sequence: $$\|F-S_n\|\downarrow0$$ This way one obtains a dominant: $$\|F-S_n\|\leq\|F-S_0\|:\quad\int\|F-S_0\|\mathrm{d}\mu\leq\int\|F\|\mathrm{d}\mu+\int\|S_0\|\mathrm{d}\mu<\infty+\infty$$ Thus by dominated convergence and linearity: $$\int F\mathrm{d}\mu-\int S_n\mathrm{d}\mu=\int(F-S_n)\mathrm{d}\mu\to0$$ (For more details see: Bochner Integral: Integrability)


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