Changing open and compact sets in regular measure definition
$begingroup$
The definition of outer regular and inner regular measures:
$begin{equation}
mu(E) = inf_{U textrm{ open}} {mu(U): U supseteq E}
end{equation}
$
$begin{equation}
mu(E) = sup_{U textrm{ compact}} {mu(U): U subseteq E}.
end{equation}$
What happened if I change open and compact sets in these definitions?
Example: Let $mu1$ be a Lebesgue measure on $mathbb{R^1}$.
Only open subset of $E1 = [0,1] cap mathbb{Q}$ is {$emptyset$}, maximal closed set is $[0,1]$. The set $E2 = [0,1]setminus mathbb{Q}$.
Both have the same properties - Only open is {$emptyset$} and max. closed set is $[0,1]$. But $mu1 = 0$ and $mu2 = 1$.
How can I interpret this result? Is there a theorem?
measure-theory
asked Dec 17 '18 at 14:46
MetsoMetso
1549
$endgroup$
add a comment |
$begingroup$
The definition of outer regular and inner regular measures:
$begin{equation}
mu(E) = inf_{U textrm{ open}} {mu(U): U supseteq E}
end{equation}
$
$begin{equation}
mu(E) = sup_{U textrm{ compact}} {mu(U): U subseteq E}.
end{equation}$
What happened if I change open and compact sets in these definitions?
Example: Let $mu1$ be a Lebesgue measure on $mathbb{R^1}$.
Only open subset of $E1 = [0,1] cap mathbb{Q}$ is {$emptyset$}, maximal closed set is $[0,1]$. The set $E2 = [0,1]setminus mathbb{Q}$.
Both have the same properties - Only open is {$emptyset$} and max. closed set is $[0,1]$. But $mu1 = 0$ and $mu2 = 1$.
How can I interpret this result? Is there a theorem?
measure-theory
asked Dec 17 '18 at 14:46
MetsoMetso
1549
$endgroup$
$begingroup$
If $E$ is unbounded and $mathbb R$ has the standard topology, your new definition won't make sense
$endgroup$
– Matematleta
Dec 17 '18 at 16:09
add a comment |
$begingroup$
The definition of outer regular and inner regular measures:
$begin{equation}
mu(E) = inf_{U textrm{ open}} {mu(U): U supseteq E}
end{equation}
$
$begin{equation}
mu(E) = sup_{U textrm{ compact}} {mu(U): U subseteq E}.
end{equation}$
What happened if I change open and compact sets in these definitions?
Example: Let $mu1$ be a Lebesgue measure on $mathbb{R^1}$.
Only open subset of $E1 = [0,1] cap mathbb{Q}$ is {$emptyset$}, maximal closed set is $[0,1]$. The set $E2 = [0,1]setminus mathbb{Q}$.
Both have the same properties - Only open is {$emptyset$} and max. closed set is $[0,1]$. But $mu1 = 0$ and $mu2 = 1$.
How can I interpret this result? Is there a theorem?
measure-theory
asked Dec 17 '18 at 14:46
MetsoMetso
1549
$endgroup$
The definition of outer regular and inner regular measures:
$begin{equation}
mu(E) = inf_{U textrm{ open}} {mu(U): U supseteq E}
end{equation}
$
$begin{equation}
mu(E) = sup_{U textrm{ compact}} {mu(U): U subseteq E}.
end{equation}$
What happened if I change open and compact sets in these definitions?
Example: Let $mu1$ be a Lebesgue measure on $mathbb{R^1}$.
Only open subset of $E1 = [0,1] cap mathbb{Q}$ is {$emptyset$}, maximal closed set is $[0,1]$. The set $E2 = [0,1]setminus mathbb{Q}$.
Both have the same properties - Only open is {$emptyset$} and max. closed set is $[0,1]$. But $mu1 = 0$ and $mu2 = 1$.
How can I interpret this result? Is there a theorem?
measure-theory
measure-theory
asked Dec 17 '18 at 14:46
MetsoMetso
1549
asked Dec 17 '18 at 14:46
MetsoMetso
1549
asked Dec 17 '18 at 14:46
MetsoMetso
1549
asked Dec 17 '18 at 14:46
MetsoMetso
1549
asked Dec 17 '18 at 14:46
MetsoMetso
1549
1549
$begingroup$
If $E$ is unbounded and $mathbb R$ has the standard topology, your new definition won't make sense
$endgroup$
– Matematleta
Dec 17 '18 at 16:09
add a comment |
$begingroup$
If $E$ is unbounded and $mathbb R$ has the standard topology, your new definition won't make sense
$endgroup$
– Matematleta
Dec 17 '18 at 16:09
$begingroup$
If $E$ is unbounded and $mathbb R$ has the standard topology, your new definition won't make sense
$endgroup$
– Matematleta
Dec 17 '18 at 16:09
$begingroup$
If $E$ is unbounded and $mathbb R$ has the standard topology, your new definition won't make sense
$endgroup$
– Matematleta
Dec 17 '18 at 16:09
add a comment |
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$begingroup$
If $E$ is unbounded and $mathbb R$ has the standard topology, your new definition won't make sense
$endgroup$
– Matematleta
Dec 17 '18 at 16:09