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 $({s}_{n})$. I know that if

${\int}_{\mathrm{\Omega}}||f(\omega )||\phantom{\rule{thickmathspace}{0ex}}d\mu (\omega )<\mathrm{\infty}\phantom{\rule{1em}{0ex}}(\ast )$

then one can show that f is Bochner-integrable, i.e. there is even a Bochner-integrable sequence of simple functions $({\stackrel{~}{s}}_{n})$ such that

$\underset{n\to \mathrm{\infty}}{lim}\phantom{\rule{thickmathspace}{0ex}}{\int}_{\mathrm{\Omega}}||f(\omega )-{\stackrel{~}{s}}_{n}(\omega )||\phantom{\rule{thickmathspace}{0ex}}d\mu (\omega )\phantom{\rule{mediummathspace}{0ex}}=\phantom{\rule{mediummathspace}{0ex}}0\phantom{\rule{1em}{0ex}}(\ast \ast )$

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 $(\ast )$, the set

$A:=\{f\ne 0\}=\bigcup _{n=1}^{\mathrm{\infty}}\phantom{\rule{thickmathspace}{0ex}}\underset{=:{A}_{n}}{\underset{\u23df}{\{||f||>\frac{1}{n}\}}}$

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 $(\ast \ast )$ anymore...any tips are much appreciated.

${\int}_{\mathrm{\Omega}}||f(\omega )||\phantom{\rule{thickmathspace}{0ex}}d\mu (\omega )<\mathrm{\infty}\phantom{\rule{1em}{0ex}}(\ast )$

then one can show that f is Bochner-integrable, i.e. there is even a Bochner-integrable sequence of simple functions $({\stackrel{~}{s}}_{n})$ such that

$\underset{n\to \mathrm{\infty}}{lim}\phantom{\rule{thickmathspace}{0ex}}{\int}_{\mathrm{\Omega}}||f(\omega )-{\stackrel{~}{s}}_{n}(\omega )||\phantom{\rule{thickmathspace}{0ex}}d\mu (\omega )\phantom{\rule{mediummathspace}{0ex}}=\phantom{\rule{mediummathspace}{0ex}}0\phantom{\rule{1em}{0ex}}(\ast \ast )$

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 $(\ast )$, the set

$A:=\{f\ne 0\}=\bigcup _{n=1}^{\mathrm{\infty}}\phantom{\rule{thickmathspace}{0ex}}\underset{=:{A}_{n}}{\underset{\u23df}{\{||f||>\frac{1}{n}\}}}$

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 $(\ast \ast )$ anymore...any tips are much appreciated.