kolutastmr

2022-07-12

"Let ${\mu }_{0}\left(A\right)=\mu \left(A\cup {A}_{0}\right),\phantom{\rule{thinmathspace}{0ex}}A\in \mathcal{F}$ be a measure. Show that if
$\int f\phantom{\rule{thinmathspace}{0ex}}d\mu$
exists, then
${\int }_{{A}_{0}}f\phantom{\rule{thinmathspace}{0ex}}d\mu =\int f\phantom{\rule{thinmathspace}{0ex}}{d}_{{\mu }_{0}}$"
However, I think there's a typo in this exercise. ${\mu }_{0}$ isn't even a measure when $\mu \left({A}_{0}\right)\ne 0$, since ${\mu }_{0}\left(\varnothing \right)=\mu \left({A}_{0}\right)$. If it were ${\mu }_{0}\left(A\right)=\mu \left(A\cap {A}_{0}\right)$, which I think that's what the author meant, I could prove it in the following way:
For simple functions,
${\int }_{{A}_{0}}f\phantom{\rule{thinmathspace}{0ex}}d\mu =\int f{\mathbf{1}}_{{A}_{0}}\phantom{\rule{thinmathspace}{0ex}}d\mu =\sum _{i=1}^{n}{f}_{i}{\mathbf{1}}_{{A}_{0}}\mu \left({A}_{i}\right)=\sum _{i=1}^{n}{f}_{i}\mu \left({A}_{i}\cap {A}_{0}\right)=\sum _{i=1}^{n}{f}_{i}{\mu }_{0}\left({A}_{i}\right)=\int f\phantom{\rule{thinmathspace}{0ex}}d{\mu }_{0}$
For a non-negative measurable function $f$, there's a non-decreasing sequence of simple functions $\left({s}_{n}{\right)}_{n\in \mathbb{N}}$ such that ${s}_{n}\to f$, then
${\int }_{{A}_{0}}f\phantom{\rule{thinmathspace}{0ex}}d\mu =\int f{\mathbf{1}}_{{A}_{0}}\phantom{\rule{thinmathspace}{0ex}}d\mu =\underset{n\to \mathrm{\infty }}{lim}\int {s}_{n}{\mathbf{1}}_{{A}_{0}}\phantom{\rule{thinmathspace}{0ex}}d\mu =\underset{n\to \mathrm{\infty }}{lim}\int {s}_{n}\phantom{\rule{thinmathspace}{0ex}}d{\mu }_{0}=\int f\phantom{\rule{thinmathspace}{0ex}}d{\mu }_{0}$
And finally, for any measurable function, we just use $f={f}^{+}-{f}^{-}$. So, which one is it, $\mu \left(A\cup {A}_{0}\right)$ or $\mu \left(A\cap {A}_{0}\right)$?

Ashley Parks

Expert

Your proof looks good (with some minor typo). I think you are right that it should be ${\mu }_{0}\left(A\right)=\mu \left(A\cap {A}_{0}\right)$.