# How to prove this condition for Bochner integrability on a general measure space?

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.

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.