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 $({s}_{n})$. 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.

Schetterai

Beginner2022-06-25Added 25 answers

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 ${\int}_{{A}_{n}}\Vert f{\chi}_{{A}_{n}}-{t}_{n}\Vert d\mu <\frac{1}{n}$. Let ${s}_{n}={t}_{n}$ on An and 0 outside. Then ${s}_{n}$ is a simple function on $\mathrm{\Omega}$ and $\int \Vert f-{s}_{n}\Vert d\mu \le {\int}_{{A}_{n}}\Vert f{\chi}_{{A}_{n}}-{t}_{n}\Vert +{\int}_{\mathrm{\Omega}\setminus {A}_{n}}\Vert f\Vert d\mu $. Now ${\int}_{\mathrm{\Omega}\setminus {A}_{n}}\Vert f\Vert d\mu \to 0$ because $\underset{n}{lim}{\int}_{{A}_{n}}\Vert f\Vert d\mu =\int \Vert f{\chi}_{\{f\ne 0\}}\Vert d\mu (\equiv \int \Vert f\Vert d\mu )$ by Monotone ConvegenceTheorem.

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