Let $f: \Omega \rightarrow B$ be Bochner-measurable, i.e. the point-wise limit of a sequence of simple (i.e. countably-valued measurable) functions $(s_n)$. I know that if
$$ \int_{\Omega} ||f(\omega)|| \; d\mu(\omega) < \infty \quad (*)$$
then one can show that $f$ is Bochner-integrable, i.e. there is even a Bochner-integrable sequence of simple functions $(\tilde{s}_n)$ such that
$$ \lim_{n \rightarrow \infty} \; \int_{\Omega} ||f(\omega)-\tilde{s}_n(\omega)|| \; d\mu(\omega) \:=\: 0 \quad (**)$$
I am unsure, however, how to derive this sequence when the measure space $\Omega$ is not finite. Apparently one is supposed to make use of the fact that due to $(*)$, the set
$$ A := \{ f \neq 0 \} = \bigcup_{n=1}^{\infty} \; \underbrace{\{ ||f|| > \frac{1}{n} \}}_{=: A_n}$$
is $\sigma$-finite, as each $A_n$ has finite measure. But how to proceed? Setting $\tilde{s}_n = 1_{A} s_n$ would not be enough to make the functions Bochner-integrable, as the entire set $A$ might still have infinite measure. Setting $\tilde{s}_n = 1_{A_n} s_n$ would do it and maintain pointwise convergence, but then I don't know how to show the convergence in $(**)$ anymore…any tips are much appreciated.
Best Answer
Think of $A_n$ as a measure space with the restriction of the $\sigma-$ algebra on $\Omega$ and the restriction of the measure $\mu$. Since you already know the result for finite measure space you can find a simple function $t_n$ on this space such that $\int_{A_n} \|f\chi_{A_n}-t_n\|d\mu <\frac 1n$. Let $s_n=t_n$ on $A_n$ and $0$ outside. Then $s_n $ is a simple function on $\Omega$ and $\int \|f-s_n\|d\mu \leq \int_{A_n} \|f\chi_{A_n}-t_n\|+\int_{\Omega \setminus A_n} \|f\|d\mu$. Now $\int_{\Omega \setminus A_n} \|f\|d\mu \to 0$ because $\lim_n \int_{A_n} \|f\|d\mu =\int \|f\chi_{\{f \neq 0\}}\|d\mu (\equiv \int \|f\|d\mu)$ by Monotone ConvegenceTheorem.