Mixing
How to prove that $L(I)=m(I setminus A)+m(A)$?
$begingroup$
Let $A$ be any subset of $mathbb R$ and let $I$ be any compact interval of $mathbb R$ containing $A$
Then I need to prove that
$$L(I)=m(I setminus A)+m(A)$$ where $L(I)$ denotes the length of $I$ and $m(A)$ denotes the outer measure of $A$
All I know regarding this problem is, if $F$ is collection of subsets of $mathbb R$ such that $A$ is in $F$ if $m(Acup B)=m(A)+m(B)$, where $B$ is any subset of $mathbb R$ such that $Acap B = varnothing$, then this collection contains all intervals.
measure-theory lebesgue-measure
edited Jan 16 at 20:00
Cameron Williams
22.4k43680
asked Jan 16 at 18:59
ArshdeepArshdeep
434
$endgroup$
add a comment |
$begingroup$
Let $A$ be any subset of $mathbb R$ and let $I$ be any compact interval of $mathbb R$ containing $A$
Then I need to prove that
$$L(I)=m(I setminus A)+m(A)$$ where $L(I)$ denotes the length of $I$ and $m(A)$ denotes the outer measure of $A$
All I know regarding this problem is, if $F$ is collection of subsets of $mathbb R$ such that $A$ is in $F$ if $m(Acup B)=m(A)+m(B)$, where $B$ is any subset of $mathbb R$ such that $Acap B = varnothing$, then this collection contains all intervals.
measure-theory lebesgue-measure
edited Jan 16 at 20:00
Cameron Williams
22.4k43680
asked Jan 16 at 18:59
ArshdeepArshdeep
434
$endgroup$
$begingroup$
Since the '' character is used as a delimiter for $LaTeX$ commands, an empty '' does nothing but put a blank space between characters. If you want a backslash to appear in math mode, do setminus. It's a very common mistake to make, but a mistake that is easily avoided.
$endgroup$
– Cameron Williams
Jan 16 at 20:01
$begingroup$
This is false. Assuming the axiom of choice, there is a subset of the unit interval such that both it and its complement have outer measure $1$.
$endgroup$
– Wojowu
Jan 16 at 20:03
$begingroup$
In fact, if I'm not mistaken, this statement is true iff $A$ is a measurable set.
$endgroup$
– Wojowu
Jan 16 at 20:08
add a comment |
$begingroup$
Let $A$ be any subset of $mathbb R$ and let $I$ be any compact interval of $mathbb R$ containing $A$
Then I need to prove that
$$L(I)=m(I setminus A)+m(A)$$ where $L(I)$ denotes the length of $I$ and $m(A)$ denotes the outer measure of $A$
All I know regarding this problem is, if $F$ is collection of subsets of $mathbb R$ such that $A$ is in $F$ if $m(Acup B)=m(A)+m(B)$, where $B$ is any subset of $mathbb R$ such that $Acap B = varnothing$, then this collection contains all intervals.
measure-theory lebesgue-measure
edited Jan 16 at 20:00
Cameron Williams
22.4k43680
asked Jan 16 at 18:59
ArshdeepArshdeep
434
$endgroup$
Let $A$ be any subset of $mathbb R$ and let $I$ be any compact interval of $mathbb R$ containing $A$
Then I need to prove that
$$L(I)=m(I setminus A)+m(A)$$ where $L(I)$ denotes the length of $I$ and $m(A)$ denotes the outer measure of $A$
All I know regarding this problem is, if $F$ is collection of subsets of $mathbb R$ such that $A$ is in $F$ if $m(Acup B)=m(A)+m(B)$, where $B$ is any subset of $mathbb R$ such that $Acap B = varnothing$, then this collection contains all intervals.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
edited Jan 16 at 20:00
Cameron Williams
22.4k43680
asked Jan 16 at 18:59
ArshdeepArshdeep
434
edited Jan 16 at 20:00
Cameron Williams
22.4k43680
asked Jan 16 at 18:59
ArshdeepArshdeep
434
edited Jan 16 at 20:00
Cameron Williams
22.4k43680
edited Jan 16 at 20:00
Cameron Williams
22.4k43680
edited Jan 16 at 20:00
Cameron Williams
22.4k43680
22.4k43680
asked Jan 16 at 18:59
ArshdeepArshdeep
434
asked Jan 16 at 18:59
ArshdeepArshdeep
434
asked Jan 16 at 18:59
ArshdeepArshdeep
434
434
$begingroup$
Since the '' character is used as a delimiter for $LaTeX$ commands, an empty '' does nothing but put a blank space between characters. If you want a backslash to appear in math mode, do setminus. It's a very common mistake to make, but a mistake that is easily avoided.
$endgroup$
– Cameron Williams
Jan 16 at 20:01
$begingroup$
This is false. Assuming the axiom of choice, there is a subset of the unit interval such that both it and its complement have outer measure $1$.
$endgroup$
– Wojowu
Jan 16 at 20:03
$begingroup$
In fact, if I'm not mistaken, this statement is true iff $A$ is a measurable set.
$endgroup$
– Wojowu
Jan 16 at 20:08
add a comment |
$begingroup$
Since the '' character is used as a delimiter for $LaTeX$ commands, an empty '' does nothing but put a blank space between characters. If you want a backslash to appear in math mode, do setminus. It's a very common mistake to make, but a mistake that is easily avoided.
$endgroup$
– Cameron Williams
Jan 16 at 20:01
$begingroup$
This is false. Assuming the axiom of choice, there is a subset of the unit interval such that both it and its complement have outer measure $1$.
$endgroup$
– Wojowu
Jan 16 at 20:03
$begingroup$
In fact, if I'm not mistaken, this statement is true iff $A$ is a measurable set.
$endgroup$
– Wojowu
Jan 16 at 20:08
$begingroup$
Since the '' character is used as a delimiter for $LaTeX$ commands, an empty '' does nothing but put a blank space between characters. If you want a backslash to appear in math mode, do setminus. It's a very common mistake to make, but a mistake that is easily avoided.
$endgroup$
– Cameron Williams
Jan 16 at 20:01
$begingroup$
Since the '' character is used as a delimiter for $LaTeX$ commands, an empty '' does nothing but put a blank space between characters. If you want a backslash to appear in math mode, do setminus. It's a very common mistake to make, but a mistake that is easily avoided.
$endgroup$
– Cameron Williams
Jan 16 at 20:01
$begingroup$
This is false. Assuming the axiom of choice, there is a subset of the unit interval such that both it and its complement have outer measure $1$.
$endgroup$
– Wojowu
Jan 16 at 20:03
$begingroup$
This is false. Assuming the axiom of choice, there is a subset of the unit interval such that both it and its complement have outer measure $1$.
$endgroup$
– Wojowu
Jan 16 at 20:03
$begingroup$
In fact, if I'm not mistaken, this statement is true iff $A$ is a measurable set.
$endgroup$
– Wojowu
Jan 16 at 20:08
$begingroup$
In fact, if I'm not mistaken, this statement is true iff $A$ is a measurable set.
$endgroup$
– Wojowu
Jan 16 at 20:08
add a comment |
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$begingroup$
Since the '' character is used as a delimiter for $LaTeX$ commands, an empty '' does nothing but put a blank space between characters. If you want a backslash to appear in math mode, do setminus. It's a very common mistake to make, but a mistake that is easily avoided.
$endgroup$
– Cameron Williams
Jan 16 at 20:01
$begingroup$
This is false. Assuming the axiom of choice, there is a subset of the unit interval such that both it and its complement have outer measure $1$.
$endgroup$
– Wojowu
Jan 16 at 20:03
$begingroup$
In fact, if I'm not mistaken, this statement is true iff $A$ is a measurable set.
$endgroup$
– Wojowu
Jan 16 at 20:08