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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.

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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.

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  • $\begingroup$ 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)! $\endgroup$ – Colescu Oct 5 '18 at 11:09

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