A semifinite measure with a locally measurable subset satisfying conditions
Definitions
A sigma algebra $(X, mathcal M)$ contains the empty set, and is closed under the formation of complements and countable unions.
A measure is a function $mu : mathcal M to [0,infty]$ tbat sends the empty set to zero and is countably additive.
A semifinite measure is one where for every $E in mathcal M$ with $mu(E) = infty$, there is $F subset E$ with $0 < mu(F) < infty$.
A set $E subseteq X$ is locally measurable iff for every $B in mathcal M$ with $mu(B) < infty$, we have $B cap E in mathcal M$.
Question
This is taken from Royden, Chapter 17, Problem 10, part (iv) (P.342), to find a semifinite measure $(X, mathcal M, mu)$ and a locally measurable set $E subset X$ that is not measurable, with:
$$sup_{F subset E, F in mathcal M} mu(F) < infty$$
My solution
Take $X = bigcup_{alpha in omega_1} {delta_alpha, varepsilon_alpha}$ with $mathcal M$ the countable/co-countable sigma algebra, and $mu({delta_alpha}) = 1$ and $mu({varepsilon_alpha})=0$, and take $E = {varepsilon_alpha mid alpha in omega_1}$.
$E$ is locally measurable: given $B in mathcal M$ with $mu(B) < infty$, $B$ cannot be co-countable, so $B$ is countable, so $B cap E$ is also countable, so $B cap E in mathcal M$.
$E$ is not measurable: $|E| = |E^c| = aleph_1$.
$sup_{F subset E, F in mathcal M} mu(F) = 0 < infty$
Is my solution correct?
measure-theory proof-verification
asked Dec 1 at 3:46
Kenny Lau
19.5k2158
add a comment |
Definitions
A sigma algebra $(X, mathcal M)$ contains the empty set, and is closed under the formation of complements and countable unions.
A measure is a function $mu : mathcal M to [0,infty]$ tbat sends the empty set to zero and is countably additive.
A semifinite measure is one where for every $E in mathcal M$ with $mu(E) = infty$, there is $F subset E$ with $0 < mu(F) < infty$.
A set $E subseteq X$ is locally measurable iff for every $B in mathcal M$ with $mu(B) < infty$, we have $B cap E in mathcal M$.
Question
This is taken from Royden, Chapter 17, Problem 10, part (iv) (P.342), to find a semifinite measure $(X, mathcal M, mu)$ and a locally measurable set $E subset X$ that is not measurable, with:
$$sup_{F subset E, F in mathcal M} mu(F) < infty$$
My solution
Take $X = bigcup_{alpha in omega_1} {delta_alpha, varepsilon_alpha}$ with $mathcal M$ the countable/co-countable sigma algebra, and $mu({delta_alpha}) = 1$ and $mu({varepsilon_alpha})=0$, and take $E = {varepsilon_alpha mid alpha in omega_1}$.
$E$ is locally measurable: given $B in mathcal M$ with $mu(B) < infty$, $B$ cannot be co-countable, so $B$ is countable, so $B cap E$ is also countable, so $B cap E in mathcal M$.
$E$ is not measurable: $|E| = |E^c| = aleph_1$.
$sup_{F subset E, F in mathcal M} mu(F) = 0 < infty$
Is my solution correct?
measure-theory proof-verification
asked Dec 1 at 3:46
Kenny Lau
19.5k2158
add a comment |
Definitions
A sigma algebra $(X, mathcal M)$ contains the empty set, and is closed under the formation of complements and countable unions.
A measure is a function $mu : mathcal M to [0,infty]$ tbat sends the empty set to zero and is countably additive.
A semifinite measure is one where for every $E in mathcal M$ with $mu(E) = infty$, there is $F subset E$ with $0 < mu(F) < infty$.
A set $E subseteq X$ is locally measurable iff for every $B in mathcal M$ with $mu(B) < infty$, we have $B cap E in mathcal M$.
Question
This is taken from Royden, Chapter 17, Problem 10, part (iv) (P.342), to find a semifinite measure $(X, mathcal M, mu)$ and a locally measurable set $E subset X$ that is not measurable, with:
$$sup_{F subset E, F in mathcal M} mu(F) < infty$$
My solution
Take $X = bigcup_{alpha in omega_1} {delta_alpha, varepsilon_alpha}$ with $mathcal M$ the countable/co-countable sigma algebra, and $mu({delta_alpha}) = 1$ and $mu({varepsilon_alpha})=0$, and take $E = {varepsilon_alpha mid alpha in omega_1}$.
$E$ is locally measurable: given $B in mathcal M$ with $mu(B) < infty$, $B$ cannot be co-countable, so $B$ is countable, so $B cap E$ is also countable, so $B cap E in mathcal M$.
$E$ is not measurable: $|E| = |E^c| = aleph_1$.
$sup_{F subset E, F in mathcal M} mu(F) = 0 < infty$
Is my solution correct?
measure-theory proof-verification
asked Dec 1 at 3:46
Kenny Lau
19.5k2158
Definitions
A sigma algebra $(X, mathcal M)$ contains the empty set, and is closed under the formation of complements and countable unions.
A measure is a function $mu : mathcal M to [0,infty]$ tbat sends the empty set to zero and is countably additive.
A semifinite measure is one where for every $E in mathcal M$ with $mu(E) = infty$, there is $F subset E$ with $0 < mu(F) < infty$.
A set $E subseteq X$ is locally measurable iff for every $B in mathcal M$ with $mu(B) < infty$, we have $B cap E in mathcal M$.
Question
This is taken from Royden, Chapter 17, Problem 10, part (iv) (P.342), to find a semifinite measure $(X, mathcal M, mu)$ and a locally measurable set $E subset X$ that is not measurable, with:
$$sup_{F subset E, F in mathcal M} mu(F) < infty$$
My solution
Take $X = bigcup_{alpha in omega_1} {delta_alpha, varepsilon_alpha}$ with $mathcal M$ the countable/co-countable sigma algebra, and $mu({delta_alpha}) = 1$ and $mu({varepsilon_alpha})=0$, and take $E = {varepsilon_alpha mid alpha in omega_1}$.
$E$ is locally measurable: given $B in mathcal M$ with $mu(B) < infty$, $B$ cannot be co-countable, so $B$ is countable, so $B cap E$ is also countable, so $B cap E in mathcal M$.
$E$ is not measurable: $|E| = |E^c| = aleph_1$.
$sup_{F subset E, F in mathcal M} mu(F) = 0 < infty$
Is my solution correct?
measure-theory proof-verification
measure-theory proof-verification
asked Dec 1 at 3:46
Kenny Lau
19.5k2158
asked Dec 1 at 3:46
Kenny Lau
19.5k2158
asked Dec 1 at 3:46
Kenny Lau
19.5k2158
asked Dec 1 at 3:46
Kenny Lau
19.5k2158
asked Dec 1 at 3:46
Kenny Lau
19.5k2158
19.5k2158
add a comment |
add a comment |
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