If $m^*(E)=0$, prove that $m^*(E_n)=0$ for each $n$
Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$.
Let $E=bigcup_{n=1}^infty E_n$. Suppose $m^*(E)=0$. Prove that $m^*(E_n)=0$ for every $n$.
My attempt:
Since $m^*(E)=0$, $E$ is a countable set. This means each of the $E_n$ is also countable and so $m^*(E_n)=0$ for all $n$.
QED
Is this correct?
real-analysis proof-verification
asked Dec 3 '18 at 0:04
Thomas
730416
add a comment |
Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$.
Let $E=bigcup_{n=1}^infty E_n$. Suppose $m^*(E)=0$. Prove that $m^*(E_n)=0$ for every $n$.
My attempt:
Since $m^*(E)=0$, $E$ is a countable set. This means each of the $E_n$ is also countable and so $m^*(E_n)=0$ for all $n$.
QED
Is this correct?
real-analysis proof-verification
asked Dec 3 '18 at 0:04
Thomas
730416
3
Not at all correct. Many uncountable sets have outer measure zero, e.g. many Cantor sets.
– T. Bongers
Dec 3 '18 at 0:10
add a comment |
Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$.
Let $E=bigcup_{n=1}^infty E_n$. Suppose $m^*(E)=0$. Prove that $m^*(E_n)=0$ for every $n$.
My attempt:
Since $m^*(E)=0$, $E$ is a countable set. This means each of the $E_n$ is also countable and so $m^*(E_n)=0$ for all $n$.
QED
Is this correct?
real-analysis proof-verification
asked Dec 3 '18 at 0:04
Thomas
730416
Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$.
Let $E=bigcup_{n=1}^infty E_n$. Suppose $m^*(E)=0$. Prove that $m^*(E_n)=0$ for every $n$.
My attempt:
Since $m^*(E)=0$, $E$ is a countable set. This means each of the $E_n$ is also countable and so $m^*(E_n)=0$ for all $n$.
QED
Is this correct?
real-analysis proof-verification
real-analysis proof-verification
asked Dec 3 '18 at 0:04
Thomas
730416
asked Dec 3 '18 at 0:04
Thomas
730416
asked Dec 3 '18 at 0:04
Thomas
730416
asked Dec 3 '18 at 0:04
Thomas
730416
asked Dec 3 '18 at 0:04
Thomas
730416
730416
3
Not at all correct. Many uncountable sets have outer measure zero, e.g. many Cantor sets.
– T. Bongers
Dec 3 '18 at 0:10
add a comment |
3
Not at all correct. Many uncountable sets have outer measure zero, e.g. many Cantor sets.
– T. Bongers
Dec 3 '18 at 0:10
3
3
Not at all correct. Many uncountable sets have outer measure zero, e.g. many Cantor sets.
– T. Bongers
Dec 3 '18 at 0:10
Not at all correct. Many uncountable sets have outer measure zero, e.g. many Cantor sets.
– T. Bongers
Dec 3 '18 at 0:10
add a comment |
1 Answer
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$E_n subseteq E$ for all $n$. By monotonicity, $m^*(E_n) = 0$.
answered Dec 3 '18 at 0:11
Lukas Kofler
1
add a comment |
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$E_n subseteq E$ for all $n$. By monotonicity, $m^*(E_n) = 0$.
answered Dec 3 '18 at 0:11
Lukas Kofler
1
add a comment |
$E_n subseteq E$ for all $n$. By monotonicity, $m^*(E_n) = 0$.
answered Dec 3 '18 at 0:11
Lukas Kofler
1
add a comment |
$E_n subseteq E$ for all $n$. By monotonicity, $m^*(E_n) = 0$.
answered Dec 3 '18 at 0:11
Lukas Kofler
1
$E_n subseteq E$ for all $n$. By monotonicity, $m^*(E_n) = 0$.
answered Dec 3 '18 at 0:11
Lukas Kofler
1
answered Dec 3 '18 at 0:11
Lukas Kofler
1
answered Dec 3 '18 at 0:11
Lukas Kofler
1
answered Dec 3 '18 at 0:11
Lukas Kofler
1
1
add a comment |
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3
Not at all correct. Many uncountable sets have outer measure zero, e.g. many Cantor sets.
– T. Bongers
Dec 3 '18 at 0:10