hawatajwizp

2022-06-24

Let $f:\mathrm{\Omega }\to B$ be Bochner-measurable, i.e. the point-wise limit of a sequence of simple (i.e. countably-valued measurable) functions $\left({s}_{n}\right)$. I know that if
${\int }_{\mathrm{\Omega }}||f\left(\omega \right)||\phantom{\rule{thickmathspace}{0ex}}d\mu \left(\omega \right)<\mathrm{\infty }\phantom{\rule{1em}{0ex}}\left(\ast \right)$
then one can show that f is Bochner-integrable, i.e. there is even a Bochner-integrable sequence of simple functions $\left({\stackrel{~}{s}}_{n}\right)$ such that
$\underset{n\to \mathrm{\infty }}{lim}\phantom{\rule{thickmathspace}{0ex}}{\int }_{\mathrm{\Omega }}||f\left(\omega \right)-{\stackrel{~}{s}}_{n}\left(\omega \right)||\phantom{\rule{thickmathspace}{0ex}}d\mu \left(\omega \right)\phantom{\rule{mediummathspace}{0ex}}=\phantom{\rule{mediummathspace}{0ex}}0\phantom{\rule{1em}{0ex}}\left(\ast \ast \right)$
I am unsure, however, how to derive this sequence when the measure space $\mathrm{\Omega }$ is not finite. Apparently one is supposed to make use of the fact that due to $\left(\ast \right)$, the set
$A:=\left\{f\ne 0\right\}=\bigcup _{n=1}^{\mathrm{\infty }}\phantom{\rule{thickmathspace}{0ex}}\underset{=:{A}_{n}}{\underset{⏟}{\left\{||f||>\frac{1}{n}\right\}}}$
is $\sigma$-finite, as each ${A}_{n}$ has finite measure. But how to proceed? Setting ${\stackrel{~}{s}}_{n}={1}_{A}{s}_{n}$ would not be enough to make the functions Bochner-integrable, as the entire set $A$ might still have infinite measure. Setting ${\stackrel{~}{s}}_{n}={1}_{{A}_{n}}{s}_{n}$ would do it and maintain pointwise convergence, but then I don't know how to show the convergence in $\left(\ast \ast \right)$ anymore...any tips are much appreciated.

Schetterai

Think of ${A}_{n}$ as a measure space with the restriction of the $\sigma -$−algebra on $\mathrm{\Omega }$ 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 ${\int }_{{A}_{n}}‖f{\chi }_{{A}_{n}}-{t}_{n}‖d\mu <\frac{1}{n}$. Let ${s}_{n}={t}_{n}$ on An and 0 outside. Then ${s}_{n}$ is a simple function on $\mathrm{\Omega }$ and $\int ‖f-{s}_{n}‖d\mu \le {\int }_{{A}_{n}}‖f{\chi }_{{A}_{n}}-{t}_{n}‖+{\int }_{\mathrm{\Omega }\setminus {A}_{n}}‖f‖d\mu$. Now ${\int }_{\mathrm{\Omega }\setminus {A}_{n}}‖f‖d\mu \to 0$ because $\underset{n}{lim}{\int }_{{A}_{n}}‖f‖d\mu =\int ‖f{\chi }_{\left\{f\ne 0\right\}}‖d\mu \left(\equiv \int ‖f‖d\mu \right)$ by Monotone ConvegenceTheorem.