# Proof of a criterion for Bochner integrability

The following comes from the Wikipedia article on Bochner integrals.

Theorem. If $$(X,\Sigma,\mu)$$ is a measure space, then a Bochner-measurable function $$f:X\to B$$ is Bochner integrable if and only if $$\int_X\|f\|_B\,d\mu<\infty.$$

Here are the relevant definitions.

Definition. A function $$f:X\to B$$  is called Bochner-measurable if it is equal $$\mu$$-almost everywhere to a function $$g$$ taking values in a separable subspace $$B_0$$ of $$B$$, and such that the inverse image $$g^{−1}(U)$$ of every open set $$U$$ in $$B$$ belongs to $$\Sigma$$.

Definition. A measurable function $$f:X\to B$$ is Bochner integrable if there exists a sequence of integrable simple functions $$s_n$$ such that $$\lim_{n\to\infty}\int_X\|f-s_n\|_B\,d\mu=0,$$ where the integral on the left-hand side is an ordinary Lebesgue integral.

How do I prove the theorem above? I'm following Analysis III by H. Amann & J. Escher, and this theorem is indeed proved, but under the assumption that $$\mu$$ is complete and $$\sigma$$-finite. The subtlety here is this (I'll stick to the notations and terminologies on the wikipedia article):

Theorem. $$f$$ is Bochner measurable if and only if $$f$$ is limit $$\mu$$-almost everywhere of a sequence of simple functions.

Theorem. If $$\mu$$ is $$\sigma$$-finite, $$f$$ is Bochner measurable if and only if $$f$$ is limit $$\mu$$-almost everywhere of a sequence of integrable simple functions.

The proof of the latter makes use of $$\sigma$$-finiteness to ensure that the the simple functions $$s_n$$ are indeed integrable, i.e., $$s_n^{-1}(B\setminus\{0\})$$ has finite measure.

On the other hand, I can prove this (by adapting the proof from Amann & Escher):

Theorem. If $$f$$ is limit $$\mu$$-almost everywhere of a sequence of simple functions, then there is a sequence of simple functions $$s_n$$ such that $$\lim_{n\to\infty}\int_X\|f-s_n\|_B\,d\mu=0.$$

Theorem. If $$f$$ is limit $$\mu$$-almost everywhere of a sequence of integrable simple functions, then there is a sequence of integrable simple functions $$s_n$$ such that $$\lim_{n\to\infty}\int_X\|f-s_n\|_B\,d\mu=0.$$

(The reason is that, the proof assumes there is a sequence of simple functions $$t_n\to f$$ a.e., and then constructs $$s_n$$ out of $$t_n$$. If $$t_n$$ are integrable, then $$s_n$$ are integrable by construction.)

Therefore, assuming $$\sigma$$-finiteness, we can show: $$f$$ is Bochner-measurable $$\implies$$ $$f$$ is limit $$\mu$$-almost everywhere of a sequence of integrable simple functions $$\implies$$ there is a sequence of integrable simple functions $$s_n$$ such that $$\lim_{n\to\infty}\int_X\|f-s_n\|_B\,d\mu=0.$$ However, I don't know how I should prove this without $$\sigma$$-finiteness.

The condition $$\int_X \|f\|\, d\mu < \infty$$ implies that $$\mu \{x: \|f(x)\| >\frac 1 n \} <\infty$$ for each $$n$$ and $$\{x: f(x) \neq 0\}$$ is the union of the sets $$\{x: \|f(x)\| >\frac 1 n \}$$ so it has sigma finite measure. Simply apply your arguments with $$\mu$$ restricted to this set of sigma finite measure. Since $$f$$ vanishes outside this set the remaining part of $$X$$ is not involved in this result at all.

• Thank you very much!! This is an important observation and it clarifies a great lot of things for me (which has been haunting me for days)! – Colescu Oct 5 '18 at 11:09