kolutastmr

Answered

2022-07-12

"Let ${\mu}_{0}(A)=\mu (A\cup {A}_{0}),\phantom{\rule{thinmathspace}{0ex}}A\in \mathcal{F}$ 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 $\mu ({A}_{0})\ne 0$, since ${\mu}_{0}(\varnothing )=\mu ({A}_{0})$. If it were ${\mu}_{0}(A)=\mu (A\cap {A}_{0})$, which I think that's what the author meant, I could prove it in the following way:

For simple functions,

${\int}_{{A}_{0}}f\phantom{\rule{thinmathspace}{0ex}}d\mu =\int f{\mathbf{1}}_{{A}_{0}}\phantom{\rule{thinmathspace}{0ex}}d\mu =\sum _{i=1}^{n}{f}_{i}{\mathbf{1}}_{{A}_{0}}\mu ({A}_{i})=\sum _{i=1}^{n}{f}_{i}\mu ({A}_{i}\cap {A}_{0})=\sum _{i=1}^{n}{f}_{i}{\mu}_{0}({A}_{i})=\int f\phantom{\rule{thinmathspace}{0ex}}d{\mu}_{0}$

For a non-negative measurable function $f$, there's a non-decreasing sequence of simple functions $({s}_{n}{)}_{n\in \mathbb{N}}$ 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, $\mu (A\cup {A}_{0})$ or $\mu (A\cap {A}_{0})$?

$\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 $\mu ({A}_{0})\ne 0$, since ${\mu}_{0}(\varnothing )=\mu ({A}_{0})$. If it were ${\mu}_{0}(A)=\mu (A\cap {A}_{0})$, which I think that's what the author meant, I could prove it in the following way:

For simple functions,

${\int}_{{A}_{0}}f\phantom{\rule{thinmathspace}{0ex}}d\mu =\int f{\mathbf{1}}_{{A}_{0}}\phantom{\rule{thinmathspace}{0ex}}d\mu =\sum _{i=1}^{n}{f}_{i}{\mathbf{1}}_{{A}_{0}}\mu ({A}_{i})=\sum _{i=1}^{n}{f}_{i}\mu ({A}_{i}\cap {A}_{0})=\sum _{i=1}^{n}{f}_{i}{\mu}_{0}({A}_{i})=\int f\phantom{\rule{thinmathspace}{0ex}}d{\mu}_{0}$

For a non-negative measurable function $f$, there's a non-decreasing sequence of simple functions $({s}_{n}{)}_{n\in \mathbb{N}}$ 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, $\mu (A\cup {A}_{0})$ or $\mu (A\cap {A}_{0})$?

Answer & Explanation

Ashley Parks

Expert

2022-07-13Added 11 answers

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

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