There is an open sequence such that its intersection covers a subset of Re and its lebesgue measure is equal...
$begingroup$
Show that for all $A subset Re$ there is a sequence of open sets $O_n$ such that $A subset cap_n 0_n$ and $ mu(cap_n 0_n)= mu^* (A)$
If A is just an whatever interval (a,b), I see that it is enough to take the sequence $O_n=(a- frac{1}{n} , b+ frac{1}{n})$. The same if it is a single set. It also works for unions of intervals e.g. $A= (a,b) cup (c,d)$ ,then $O_n=(a-frac {1}{n}, b+frac{1}{n}) cup (c- frac{1}{n}, d+frac{1}{n})$. Is there something I am missing? How can I prove it more generally?
measure-theory
asked Dec 6 '18 at 17:16
Manos LouManos Lou
62
$endgroup$
add a comment |
$begingroup$
Show that for all $A subset Re$ there is a sequence of open sets $O_n$ such that $A subset cap_n 0_n$ and $ mu(cap_n 0_n)= mu^* (A)$
If A is just an whatever interval (a,b), I see that it is enough to take the sequence $O_n=(a- frac{1}{n} , b+ frac{1}{n})$. The same if it is a single set. It also works for unions of intervals e.g. $A= (a,b) cup (c,d)$ ,then $O_n=(a-frac {1}{n}, b+frac{1}{n}) cup (c- frac{1}{n}, d+frac{1}{n})$. Is there something I am missing? How can I prove it more generally?
measure-theory
asked Dec 6 '18 at 17:16
Manos LouManos Lou
62
$endgroup$
add a comment |
$begingroup$
Show that for all $A subset Re$ there is a sequence of open sets $O_n$ such that $A subset cap_n 0_n$ and $ mu(cap_n 0_n)= mu^* (A)$
If A is just an whatever interval (a,b), I see that it is enough to take the sequence $O_n=(a- frac{1}{n} , b+ frac{1}{n})$. The same if it is a single set. It also works for unions of intervals e.g. $A= (a,b) cup (c,d)$ ,then $O_n=(a-frac {1}{n}, b+frac{1}{n}) cup (c- frac{1}{n}, d+frac{1}{n})$. Is there something I am missing? How can I prove it more generally?
measure-theory
asked Dec 6 '18 at 17:16
Manos LouManos Lou
62
$endgroup$
Show that for all $A subset Re$ there is a sequence of open sets $O_n$ such that $A subset cap_n 0_n$ and $ mu(cap_n 0_n)= mu^* (A)$
If A is just an whatever interval (a,b), I see that it is enough to take the sequence $O_n=(a- frac{1}{n} , b+ frac{1}{n})$. The same if it is a single set. It also works for unions of intervals e.g. $A= (a,b) cup (c,d)$ ,then $O_n=(a-frac {1}{n}, b+frac{1}{n}) cup (c- frac{1}{n}, d+frac{1}{n})$. Is there something I am missing? How can I prove it more generally?
measure-theory
measure-theory
asked Dec 6 '18 at 17:16
Manos LouManos Lou
62
asked Dec 6 '18 at 17:16
Manos LouManos Lou
62
asked Dec 6 '18 at 17:16
Manos LouManos Lou
62
asked Dec 6 '18 at 17:16
Manos LouManos Lou
62
asked Dec 6 '18 at 17:16
Manos LouManos Lou
62
62
add a comment |
add a comment |
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