What is the most elegant known proof of $m^*(A) leq sum_{n} m^*(A_n)$ when $A = bigcup_n A_n$?












1












$begingroup$


Let $A = bigcup_{n in I} A_n subset Bbb{R}^k$ where $I$ is an arbitrary index set. Define the Lebesgue outer measure by




$m^*(A) := inf { sum_{n} text{vol}(I_n) : I_n, n geq 1$ are each any type of interval and $A subset bigcup_n I_n }$.




Then how can we prove $m^*(A) leq sum_n m^*(A_n)$ elegantly. The proof in my book is kind of hand-wavy and very complicated for something intuitively obvious.



Also this is the first property of such objects other than $m^*$ is monotonic: $A subset B implies m^*(A) leq m^*(B)$. First properties should be easily proven or something is wrong with the proof technique! It has to be fixed.



So I am on the search for an elegant proof. Maybe Galois connections?



The idea is that if I am successful, I can apply the proof technique to other such over-complicated examples I encounter.






measure-theory lebesgue-measure outer-measure







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that, for uncountable $I$, the inequality is not true.
    $endgroup$
    – Martin Argerami
    Jan 12 at 4:33
















1












$begingroup$


Let $A = bigcup_{n in I} A_n subset Bbb{R}^k$ where $I$ is an arbitrary index set. Define the Lebesgue outer measure by




$m^*(A) := inf { sum_{n} text{vol}(I_n) : I_n, n geq 1$ are each any type of interval and $A subset bigcup_n I_n }$.




Then how can we prove $m^*(A) leq sum_n m^*(A_n)$ elegantly. The proof in my book is kind of hand-wavy and very complicated for something intuitively obvious.



Also this is the first property of such objects other than $m^*$ is monotonic: $A subset B implies m^*(A) leq m^*(B)$. First properties should be easily proven or something is wrong with the proof technique! It has to be fixed.



So I am on the search for an elegant proof. Maybe Galois connections?



The idea is that if I am successful, I can apply the proof technique to other such over-complicated examples I encounter.






measure-theory lebesgue-measure outer-measure







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that, for uncountable $I$, the inequality is not true.
    $endgroup$
    – Martin Argerami
    Jan 12 at 4:33














1












1








1


0



$begingroup$


Let $A = bigcup_{n in I} A_n subset Bbb{R}^k$ where $I$ is an arbitrary index set. Define the Lebesgue outer measure by




$m^*(A) := inf { sum_{n} text{vol}(I_n) : I_n, n geq 1$ are each any type of interval and $A subset bigcup_n I_n }$.




Then how can we prove $m^*(A) leq sum_n m^*(A_n)$ elegantly. The proof in my book is kind of hand-wavy and very complicated for something intuitively obvious.



Also this is the first property of such objects other than $m^*$ is monotonic: $A subset B implies m^*(A) leq m^*(B)$. First properties should be easily proven or something is wrong with the proof technique! It has to be fixed.



So I am on the search for an elegant proof. Maybe Galois connections?



The idea is that if I am successful, I can apply the proof technique to other such over-complicated examples I encounter.






measure-theory lebesgue-measure outer-measure







share|cite|improve this question











$endgroup$




Let $A = bigcup_{n in I} A_n subset Bbb{R}^k$ where $I$ is an arbitrary index set. Define the Lebesgue outer measure by




$m^*(A) := inf { sum_{n} text{vol}(I_n) : I_n, n geq 1$ are each any type of interval and $A subset bigcup_n I_n }$.




Then how can we prove $m^*(A) leq sum_n m^*(A_n)$ elegantly. The proof in my book is kind of hand-wavy and very complicated for something intuitively obvious.



Also this is the first property of such objects other than $m^*$ is monotonic: $A subset B implies m^*(A) leq m^*(B)$. First properties should be easily proven or something is wrong with the proof technique! It has to be fixed.



So I am on the search for an elegant proof. Maybe Galois connections?



The idea is that if I am successful, I can apply the proof technique to other such over-complicated examples I encounter.






measure-theory lebesgue-measure outer-measure




measure-theory lebesgue-measure outer-measure






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 2:23









David C. Ullrich

60k43994




60k43994










asked Jan 12 at 1:41









Hermit with AdjointHermit with Adjoint

9,11552458




9,11552458








  • 1




    $begingroup$
    Note that, for uncountable $I$, the inequality is not true.
    $endgroup$
    – Martin Argerami
    Jan 12 at 4:33














  • 1




    $begingroup$
    Note that, for uncountable $I$, the inequality is not true.
    $endgroup$
    – Martin Argerami
    Jan 12 at 4:33








1




1




$begingroup$
Note that, for uncountable $I$, the inequality is not true.
$endgroup$
– Martin Argerami
Jan 12 at 4:33




$begingroup$
Note that, for uncountable $I$, the inequality is not true.
$endgroup$
– Martin Argerami
Jan 12 at 4:33










1 Answer
1






active

oldest

votes


















3












$begingroup$

(This answer assumes that $I$ is countable)



This is a case for the $epsilon/2^n$ trick, in combination with the "give yourself an epsilon of room" trick. Once we've internalized these tricks the proof seems straightforward.



Let $epsilon > 0$. For each positive integer $n$, let ${I_{nj}}$ be a countable collection of intervals such that $A_n subset cup_j I_{nj}$ and
$$sum_{j} |I_{nj}| leq m^*(A_n) + frac{epsilon}{2^n}.
$$
Then



$$
A subset cup_{n,j} I_{nj}
$$
and



begin{align}
m^*(A) &leq sum_{n,j} |I_{nj}| \
&= sum_{n=1}^infty sum_{j=1}^infty |I_{nj}| \
&leq sum_{n=1}^infty m^*(A_n) + frac{epsilon}{2^n} \
&= sum_{n=1}^infty m^*(A_n) + underbrace{sum_{n=1}^infty frac{epsilon}{2^n}}_epsilon.
end{align}



This shows that
$$
m^*(A) leq sum_{n=1}^infty m^*(A_n) + epsilon
$$
for any $epsilon > 0$. It follows that
$$
m^*(A) leq sum_{n=1}^infty m^*(A_n).
$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You should note that, in the question, the union was not necessarily assumed to be countable.
    $endgroup$
    – Theo Bendit
    Jan 12 at 2:11






  • 1




    $begingroup$
    @DavidC.Ullrich I thought about that, but sets with zero outer measure need not be empty. I don't see how the $varepsilon / 2^n$ trick works in those circumstances.
    $endgroup$
    – Theo Bendit
    Jan 12 at 2:47






  • 1




    $begingroup$
    When the union is uncountable the inequality is not true, so I'm not sure what you expect to prove.
    $endgroup$
    – Martin Argerami
    Jan 12 at 3:19






  • 3




    $begingroup$
    Yes. $[0,1]=bigcup_{tin[0,1]}{t}$.
    $endgroup$
    – Martin Argerami
    Jan 12 at 3:28






  • 1




    $begingroup$
    @MartinArgerami Ugh, obviously. :-(
    $endgroup$
    – Theo Bendit
    Jan 12 at 4:11











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

(This answer assumes that $I$ is countable)



This is a case for the $epsilon/2^n$ trick, in combination with the "give yourself an epsilon of room" trick. Once we've internalized these tricks the proof seems straightforward.



Let $epsilon > 0$. For each positive integer $n$, let ${I_{nj}}$ be a countable collection of intervals such that $A_n subset cup_j I_{nj}$ and
$$sum_{j} |I_{nj}| leq m^*(A_n) + frac{epsilon}{2^n}.
$$
Then



$$
A subset cup_{n,j} I_{nj}
$$
and



begin{align}
m^*(A) &leq sum_{n,j} |I_{nj}| \
&= sum_{n=1}^infty sum_{j=1}^infty |I_{nj}| \
&leq sum_{n=1}^infty m^*(A_n) + frac{epsilon}{2^n} \
&= sum_{n=1}^infty m^*(A_n) + underbrace{sum_{n=1}^infty frac{epsilon}{2^n}}_epsilon.
end{align}



This shows that
$$
m^*(A) leq sum_{n=1}^infty m^*(A_n) + epsilon
$$
for any $epsilon > 0$. It follows that
$$
m^*(A) leq sum_{n=1}^infty m^*(A_n).
$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You should note that, in the question, the union was not necessarily assumed to be countable.
    $endgroup$
    – Theo Bendit
    Jan 12 at 2:11






  • 1




    $begingroup$
    @DavidC.Ullrich I thought about that, but sets with zero outer measure need not be empty. I don't see how the $varepsilon / 2^n$ trick works in those circumstances.
    $endgroup$
    – Theo Bendit
    Jan 12 at 2:47






  • 1




    $begingroup$
    When the union is uncountable the inequality is not true, so I'm not sure what you expect to prove.
    $endgroup$
    – Martin Argerami
    Jan 12 at 3:19






  • 3




    $begingroup$
    Yes. $[0,1]=bigcup_{tin[0,1]}{t}$.
    $endgroup$
    – Martin Argerami
    Jan 12 at 3:28






  • 1




    $begingroup$
    @MartinArgerami Ugh, obviously. :-(
    $endgroup$
    – Theo Bendit
    Jan 12 at 4:11
















3












$begingroup$

(This answer assumes that $I$ is countable)



This is a case for the $epsilon/2^n$ trick, in combination with the "give yourself an epsilon of room" trick. Once we've internalized these tricks the proof seems straightforward.



Let $epsilon > 0$. For each positive integer $n$, let ${I_{nj}}$ be a countable collection of intervals such that $A_n subset cup_j I_{nj}$ and
$$sum_{j} |I_{nj}| leq m^*(A_n) + frac{epsilon}{2^n}.
$$
Then



$$
A subset cup_{n,j} I_{nj}
$$
and



begin{align}
m^*(A) &leq sum_{n,j} |I_{nj}| \
&= sum_{n=1}^infty sum_{j=1}^infty |I_{nj}| \
&leq sum_{n=1}^infty m^*(A_n) + frac{epsilon}{2^n} \
&= sum_{n=1}^infty m^*(A_n) + underbrace{sum_{n=1}^infty frac{epsilon}{2^n}}_epsilon.
end{align}



This shows that
$$
m^*(A) leq sum_{n=1}^infty m^*(A_n) + epsilon
$$
for any $epsilon > 0$. It follows that
$$
m^*(A) leq sum_{n=1}^infty m^*(A_n).
$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You should note that, in the question, the union was not necessarily assumed to be countable.
    $endgroup$
    – Theo Bendit
    Jan 12 at 2:11






  • 1




    $begingroup$
    @DavidC.Ullrich I thought about that, but sets with zero outer measure need not be empty. I don't see how the $varepsilon / 2^n$ trick works in those circumstances.
    $endgroup$
    – Theo Bendit
    Jan 12 at 2:47






  • 1




    $begingroup$
    When the union is uncountable the inequality is not true, so I'm not sure what you expect to prove.
    $endgroup$
    – Martin Argerami
    Jan 12 at 3:19






  • 3




    $begingroup$
    Yes. $[0,1]=bigcup_{tin[0,1]}{t}$.
    $endgroup$
    – Martin Argerami
    Jan 12 at 3:28






  • 1




    $begingroup$
    @MartinArgerami Ugh, obviously. :-(
    $endgroup$
    – Theo Bendit
    Jan 12 at 4:11














3












3








3





$begingroup$

(This answer assumes that $I$ is countable)



This is a case for the $epsilon/2^n$ trick, in combination with the "give yourself an epsilon of room" trick. Once we've internalized these tricks the proof seems straightforward.



Let $epsilon > 0$. For each positive integer $n$, let ${I_{nj}}$ be a countable collection of intervals such that $A_n subset cup_j I_{nj}$ and
$$sum_{j} |I_{nj}| leq m^*(A_n) + frac{epsilon}{2^n}.
$$
Then



$$
A subset cup_{n,j} I_{nj}
$$
and



begin{align}
m^*(A) &leq sum_{n,j} |I_{nj}| \
&= sum_{n=1}^infty sum_{j=1}^infty |I_{nj}| \
&leq sum_{n=1}^infty m^*(A_n) + frac{epsilon}{2^n} \
&= sum_{n=1}^infty m^*(A_n) + underbrace{sum_{n=1}^infty frac{epsilon}{2^n}}_epsilon.
end{align}



This shows that
$$
m^*(A) leq sum_{n=1}^infty m^*(A_n) + epsilon
$$
for any $epsilon > 0$. It follows that
$$
m^*(A) leq sum_{n=1}^infty m^*(A_n).
$$






share|cite|improve this answer











$endgroup$



(This answer assumes that $I$ is countable)



This is a case for the $epsilon/2^n$ trick, in combination with the "give yourself an epsilon of room" trick. Once we've internalized these tricks the proof seems straightforward.



Let $epsilon > 0$. For each positive integer $n$, let ${I_{nj}}$ be a countable collection of intervals such that $A_n subset cup_j I_{nj}$ and
$$sum_{j} |I_{nj}| leq m^*(A_n) + frac{epsilon}{2^n}.
$$
Then



$$
A subset cup_{n,j} I_{nj}
$$
and



begin{align}
m^*(A) &leq sum_{n,j} |I_{nj}| \
&= sum_{n=1}^infty sum_{j=1}^infty |I_{nj}| \
&leq sum_{n=1}^infty m^*(A_n) + frac{epsilon}{2^n} \
&= sum_{n=1}^infty m^*(A_n) + underbrace{sum_{n=1}^infty frac{epsilon}{2^n}}_epsilon.
end{align}



This shows that
$$
m^*(A) leq sum_{n=1}^infty m^*(A_n) + epsilon
$$
for any $epsilon > 0$. It follows that
$$
m^*(A) leq sum_{n=1}^infty m^*(A_n).
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 12 at 3:37









Martin Argerami

126k1182180




126k1182180










answered Jan 12 at 2:08









littleOlittleO

29.6k646109




29.6k646109








  • 1




    $begingroup$
    You should note that, in the question, the union was not necessarily assumed to be countable.
    $endgroup$
    – Theo Bendit
    Jan 12 at 2:11






  • 1




    $begingroup$
    @DavidC.Ullrich I thought about that, but sets with zero outer measure need not be empty. I don't see how the $varepsilon / 2^n$ trick works in those circumstances.
    $endgroup$
    – Theo Bendit
    Jan 12 at 2:47






  • 1




    $begingroup$
    When the union is uncountable the inequality is not true, so I'm not sure what you expect to prove.
    $endgroup$
    – Martin Argerami
    Jan 12 at 3:19






  • 3




    $begingroup$
    Yes. $[0,1]=bigcup_{tin[0,1]}{t}$.
    $endgroup$
    – Martin Argerami
    Jan 12 at 3:28






  • 1




    $begingroup$
    @MartinArgerami Ugh, obviously. :-(
    $endgroup$
    – Theo Bendit
    Jan 12 at 4:11














  • 1




    $begingroup$
    You should note that, in the question, the union was not necessarily assumed to be countable.
    $endgroup$
    – Theo Bendit
    Jan 12 at 2:11






  • 1




    $begingroup$
    @DavidC.Ullrich I thought about that, but sets with zero outer measure need not be empty. I don't see how the $varepsilon / 2^n$ trick works in those circumstances.
    $endgroup$
    – Theo Bendit
    Jan 12 at 2:47






  • 1




    $begingroup$
    When the union is uncountable the inequality is not true, so I'm not sure what you expect to prove.
    $endgroup$
    – Martin Argerami
    Jan 12 at 3:19






  • 3




    $begingroup$
    Yes. $[0,1]=bigcup_{tin[0,1]}{t}$.
    $endgroup$
    – Martin Argerami
    Jan 12 at 3:28






  • 1




    $begingroup$
    @MartinArgerami Ugh, obviously. :-(
    $endgroup$
    – Theo Bendit
    Jan 12 at 4:11








1




1




$begingroup$
You should note that, in the question, the union was not necessarily assumed to be countable.
$endgroup$
– Theo Bendit
Jan 12 at 2:11




$begingroup$
You should note that, in the question, the union was not necessarily assumed to be countable.
$endgroup$
– Theo Bendit
Jan 12 at 2:11




1




1




$begingroup$
@DavidC.Ullrich I thought about that, but sets with zero outer measure need not be empty. I don't see how the $varepsilon / 2^n$ trick works in those circumstances.
$endgroup$
– Theo Bendit
Jan 12 at 2:47




$begingroup$
@DavidC.Ullrich I thought about that, but sets with zero outer measure need not be empty. I don't see how the $varepsilon / 2^n$ trick works in those circumstances.
$endgroup$
– Theo Bendit
Jan 12 at 2:47




1




1




$begingroup$
When the union is uncountable the inequality is not true, so I'm not sure what you expect to prove.
$endgroup$
– Martin Argerami
Jan 12 at 3:19




$begingroup$
When the union is uncountable the inequality is not true, so I'm not sure what you expect to prove.
$endgroup$
– Martin Argerami
Jan 12 at 3:19




3




3




$begingroup$
Yes. $[0,1]=bigcup_{tin[0,1]}{t}$.
$endgroup$
– Martin Argerami
Jan 12 at 3:28




$begingroup$
Yes. $[0,1]=bigcup_{tin[0,1]}{t}$.
$endgroup$
– Martin Argerami
Jan 12 at 3:28




1




1




$begingroup$
@MartinArgerami Ugh, obviously. :-(
$endgroup$
– Theo Bendit
Jan 12 at 4:11




$begingroup$
@MartinArgerami Ugh, obviously. :-(
$endgroup$
– Theo Bendit
Jan 12 at 4:11


















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