The measure of the union of 2 sets. The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Using the concept of $G_delta$ sets, show that the union of two measurable sets is measurable.There is a measure $mu^(2)(A_1times A_2)=mu(A_1cap A_2)$Is the measure space $(mathbbR, B(mathbbR),mu)$ complete where $mu$ is the Lesbesgue measure?A question on atoms in measure theory.Outer measure and trace sigma algebraCompletion of a measure space and Null SetsLet $Esubseteq[a,b]$ be measurable. Prove that $overline mA=overline m(Acap E) + overline m(Acap E^c)$ for any $Asubseteq[a,b]$.Completion of a measure space using null setsComplete Measure spacesσ-algebra and measure of the union of 2 sets. Based on measure theory.
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The measure of the union of 2 sets.
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Using the concept of $G_delta$ sets, show that the union of two measurable sets is measurable.There is a measure $mu^(2)(A_1times A_2)=mu(A_1cap A_2)$Is the measure space $(mathbbR, B(mathbbR),mu)$ complete where $mu$ is the Lesbesgue measure?A question on atoms in measure theory.Outer measure and trace sigma algebraCompletion of a measure space and Null SetsLet $Esubseteq[a,b]$ be measurable. Prove that $overline mA=overline m(Acap E) + overline m(Acap E^c)$ for any $Asubseteq[a,b]$.Completion of a measure space using null setsComplete Measure spacesσ-algebra and measure of the union of 2 sets. Based on measure theory.
$begingroup$
A measure space $(mathbb S,mathcal S,μ)$ is not complete. The system of all its null sets is $mathcal O$.
Let
$$ S′=A∪O:A∈ mathcal S,O∈ mathcal O.$$
The formula of the function $μ′:mathcal S′→[0,∞]$ is
$$μ′(A∪O)=μ(A).$$
Question:
(a) Show: $μ′(A_1∪O_1)=μ′(A_2∪O_2),$ if $A_1∪O_1=A_2∪O_2.$
My solution:
I am given some hints. I need to show that $A=A_1∪O_1=A_2∪O_2$ in two ways, with $A_1,A_2∈ mathcal S$ and $O_1,O_2∈ mathcal O$, then $μ(A_1)=μ(A_2)$. Then according to $μ′(A∪O)=μ(A)$ the final fact will follow.
measure-theory functions
edited Mar 25 at 11:11
Asaf Karagila♦
308k33441775
asked Mar 25 at 8:20
PhilipPhilip
917
$endgroup$
add a comment |
$begingroup$
A measure space $(mathbb S,mathcal S,μ)$ is not complete. The system of all its null sets is $mathcal O$.
Let
$$ S′=A∪O:A∈ mathcal S,O∈ mathcal O.$$
The formula of the function $μ′:mathcal S′→[0,∞]$ is
$$μ′(A∪O)=μ(A).$$
Question:
(a) Show: $μ′(A_1∪O_1)=μ′(A_2∪O_2),$ if $A_1∪O_1=A_2∪O_2.$
My solution:
I am given some hints. I need to show that $A=A_1∪O_1=A_2∪O_2$ in two ways, with $A_1,A_2∈ mathcal S$ and $O_1,O_2∈ mathcal O$, then $μ(A_1)=μ(A_2)$. Then according to $μ′(A∪O)=μ(A)$ the final fact will follow.
measure-theory functions
edited Mar 25 at 11:11
Asaf Karagila♦
308k33441775
asked Mar 25 at 8:20
PhilipPhilip
917
$endgroup$
add a comment |
$begingroup$
A measure space $(mathbb S,mathcal S,μ)$ is not complete. The system of all its null sets is $mathcal O$.
Let
$$ S′=A∪O:A∈ mathcal S,O∈ mathcal O.$$
The formula of the function $μ′:mathcal S′→[0,∞]$ is
$$μ′(A∪O)=μ(A).$$
Question:
(a) Show: $μ′(A_1∪O_1)=μ′(A_2∪O_2),$ if $A_1∪O_1=A_2∪O_2.$
My solution:
I am given some hints. I need to show that $A=A_1∪O_1=A_2∪O_2$ in two ways, with $A_1,A_2∈ mathcal S$ and $O_1,O_2∈ mathcal O$, then $μ(A_1)=μ(A_2)$. Then according to $μ′(A∪O)=μ(A)$ the final fact will follow.
measure-theory functions
edited Mar 25 at 11:11
Asaf Karagila♦
308k33441775
asked Mar 25 at 8:20
PhilipPhilip
917
$endgroup$
A measure space $(mathbb S,mathcal S,μ)$ is not complete. The system of all its null sets is $mathcal O$.
Let
$$ S′=A∪O:A∈ mathcal S,O∈ mathcal O.$$
The formula of the function $μ′:mathcal S′→[0,∞]$ is
$$μ′(A∪O)=μ(A).$$
Question:
(a) Show: $μ′(A_1∪O_1)=μ′(A_2∪O_2),$ if $A_1∪O_1=A_2∪O_2.$
My solution:
I am given some hints. I need to show that $A=A_1∪O_1=A_2∪O_2$ in two ways, with $A_1,A_2∈ mathcal S$ and $O_1,O_2∈ mathcal O$, then $μ(A_1)=μ(A_2)$. Then according to $μ′(A∪O)=μ(A)$ the final fact will follow.
measure-theory functions
measure-theory functions
edited Mar 25 at 11:11
Asaf Karagila♦
308k33441775
asked Mar 25 at 8:20
PhilipPhilip
917
edited Mar 25 at 11:11
Asaf Karagila♦
308k33441775
asked Mar 25 at 8:20
PhilipPhilip
917
edited Mar 25 at 11:11
Asaf Karagila♦
308k33441775
edited Mar 25 at 11:11
Asaf Karagila♦
308k33441775
edited Mar 25 at 11:11
Asaf Karagila♦
308k33441775
308k33441775
asked Mar 25 at 8:20
PhilipPhilip
917
asked Mar 25 at 8:20
PhilipPhilip
917
asked Mar 25 at 8:20
PhilipPhilip
917
917
add a comment |
add a comment |
1 Answer
1
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$begingroup$
$A_2 setminus A_1 subset O_1 $ so $mu (A_2 setminus A_1)=0$. Similarly, $mu (A_1 setminus A_2)=0$. Now $mu (A_1)leq mu (A_1 setminus A_2)+mu (A_2)=mu(A_2)$ and the reverse inequality is similar.
answered Mar 25 at 8:29
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
add a comment |
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$begingroup$
$A_2 setminus A_1 subset O_1 $ so $mu (A_2 setminus A_1)=0$. Similarly, $mu (A_1 setminus A_2)=0$. Now $mu (A_1)leq mu (A_1 setminus A_2)+mu (A_2)=mu(A_2)$ and the reverse inequality is similar.
answered Mar 25 at 8:29
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
add a comment |
$begingroup$
$A_2 setminus A_1 subset O_1 $ so $mu (A_2 setminus A_1)=0$. Similarly, $mu (A_1 setminus A_2)=0$. Now $mu (A_1)leq mu (A_1 setminus A_2)+mu (A_2)=mu(A_2)$ and the reverse inequality is similar.
answered Mar 25 at 8:29
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
add a comment |
$begingroup$
$A_2 setminus A_1 subset O_1 $ so $mu (A_2 setminus A_1)=0$. Similarly, $mu (A_1 setminus A_2)=0$. Now $mu (A_1)leq mu (A_1 setminus A_2)+mu (A_2)=mu(A_2)$ and the reverse inequality is similar.
answered Mar 25 at 8:29
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$A_2 setminus A_1 subset O_1 $ so $mu (A_2 setminus A_1)=0$. Similarly, $mu (A_1 setminus A_2)=0$. Now $mu (A_1)leq mu (A_1 setminus A_2)+mu (A_2)=mu(A_2)$ and the reverse inequality is similar.
answered Mar 25 at 8:29
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Mar 25 at 8:29
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Mar 25 at 8:29
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Mar 25 at 8:29
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
74.6k53270
add a comment |
add a comment |
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