Why isn't this a Jordan-Measurable setConstructing a number not in $bigcuplimits_k=1^infty (q_k-fracepsilon2^k,q_k+fracepsilon2^k)$R as a union of a zero measure set and a meager setFinding an irrational not covered in standard proof that $mu(mathbbQ cap [0,1]) = 0$Closure, Interior, and Boundary of Jordan Measurable Sets.A set of small measure in the real line with an interesting propertyMeasurability of a set in Littlewood 3rd principleJordan outer content and Lebesgue measure coincide on compact sets: Folland's proof.Why do we need $beta_1>0$ and $alpha_n<beta_n$ in this proof?Measure of complement of union of nowhere dense set with positive measure$f: mathbb R to mathbb R, fin C^1$ then $f$ perserves measurability
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Why isn't this a Jordan-Measurable set
Constructing a number not in $bigcuplimits_k=1^infty (q_k-fracepsilon2^k,q_k+fracepsilon2^k)$R as a union of a zero measure set and a meager setFinding an irrational not covered in standard proof that $mu(mathbbQ cap [0,1]) = 0$Closure, Interior, and Boundary of Jordan Measurable Sets.A set of small measure in the real line with an interesting propertyMeasurability of a set in Littlewood 3rd principleJordan outer content and Lebesgue measure coincide on compact sets: Folland's proof.Why do we need $beta_1>0$ and $alpha_n<beta_n$ in this proof?Measure of complement of union of nowhere dense set with positive measure$f: mathbb R to mathbb R, fin C^1$ then $f$ perserves measurability
$begingroup$
so I have a question. If we consider an enumeration of the rationals in $[0,1]$ $mathbbQ = q_1,q_2,...,q_n$ and we define $U_n,epsilon = ]q_n-epsilon/2^n, q_n + epsilon/2^n[$,
$U_epsilon = bigcuplimits_n=1^infty U_n,epsilon$
And this set won't be Jordan-Measurable at least when $epsilon $ is lower than $1/2$. My question is why it isn't possible to construct a sequence of elementary sets $K_N, U_N$ such that $K_N subseteq U_epsilon subseteq U_N$ and the content of $U_N-K_N$ goes to zero? Why doesn't making $K_N=bigcuplimits_n=1^N U_n,epsilon$ and $U_N = K_N bigcup [q_N,1]$ work? Why doesn't the limit tend to zero?
real-analysis measure-theory
edited Mar 17 at 12:16
Bernard
123k741117
asked Mar 17 at 11:58
Pedro SantosPedro Santos
1609
$endgroup$
add a comment |
$begingroup$
so I have a question. If we consider an enumeration of the rationals in $[0,1]$ $mathbbQ = q_1,q_2,...,q_n$ and we define $U_n,epsilon = ]q_n-epsilon/2^n, q_n + epsilon/2^n[$,
$U_epsilon = bigcuplimits_n=1^infty U_n,epsilon$
And this set won't be Jordan-Measurable at least when $epsilon $ is lower than $1/2$. My question is why it isn't possible to construct a sequence of elementary sets $K_N, U_N$ such that $K_N subseteq U_epsilon subseteq U_N$ and the content of $U_N-K_N$ goes to zero? Why doesn't making $K_N=bigcuplimits_n=1^N U_n,epsilon$ and $U_N = K_N bigcup [q_N,1]$ work? Why doesn't the limit tend to zero?
real-analysis measure-theory
edited Mar 17 at 12:16
Bernard
123k741117
asked Mar 17 at 11:58
Pedro SantosPedro Santos
1609
$endgroup$
add a comment |
$begingroup$
so I have a question. If we consider an enumeration of the rationals in $[0,1]$ $mathbbQ = q_1,q_2,...,q_n$ and we define $U_n,epsilon = ]q_n-epsilon/2^n, q_n + epsilon/2^n[$,
$U_epsilon = bigcuplimits_n=1^infty U_n,epsilon$
And this set won't be Jordan-Measurable at least when $epsilon $ is lower than $1/2$. My question is why it isn't possible to construct a sequence of elementary sets $K_N, U_N$ such that $K_N subseteq U_epsilon subseteq U_N$ and the content of $U_N-K_N$ goes to zero? Why doesn't making $K_N=bigcuplimits_n=1^N U_n,epsilon$ and $U_N = K_N bigcup [q_N,1]$ work? Why doesn't the limit tend to zero?
real-analysis measure-theory
edited Mar 17 at 12:16
Bernard
123k741117
asked Mar 17 at 11:58
Pedro SantosPedro Santos
1609
$endgroup$
so I have a question. If we consider an enumeration of the rationals in $[0,1]$ $mathbbQ = q_1,q_2,...,q_n$ and we define $U_n,epsilon = ]q_n-epsilon/2^n, q_n + epsilon/2^n[$,
$U_epsilon = bigcuplimits_n=1^infty U_n,epsilon$
And this set won't be Jordan-Measurable at least when $epsilon $ is lower than $1/2$. My question is why it isn't possible to construct a sequence of elementary sets $K_N, U_N$ such that $K_N subseteq U_epsilon subseteq U_N$ and the content of $U_N-K_N$ goes to zero? Why doesn't making $K_N=bigcuplimits_n=1^N U_n,epsilon$ and $U_N = K_N bigcup [q_N,1]$ work? Why doesn't the limit tend to zero?
real-analysis measure-theory
real-analysis measure-theory
edited Mar 17 at 12:16
Bernard
123k741117
asked Mar 17 at 11:58
Pedro SantosPedro Santos
1609
edited Mar 17 at 12:16
Bernard
123k741117
asked Mar 17 at 11:58
Pedro SantosPedro Santos
1609
edited Mar 17 at 12:16
Bernard
123k741117
edited Mar 17 at 12:16
Bernard
123k741117
edited Mar 17 at 12:16
Bernard
123k741117
123k741117
asked Mar 17 at 11:58
Pedro SantosPedro Santos
1609
asked Mar 17 at 11:58
Pedro SantosPedro Santos
1609
asked Mar 17 at 11:58
Pedro SantosPedro Santos
1609
1609
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
If $q_N to 1$, why must we have $U_epsilon subset U_N$?
answered Mar 17 at 12:39
RRLRRL
53k42573
$endgroup$
$begingroup$
Hm i see ur point , that neighborhood can surpass the interval .
$endgroup$
– Pedro Santos
Mar 17 at 12:40
$begingroup$
Also the enumeration cannot be monotonically increasing.
$endgroup$
– RRL
Mar 17 at 12:42
$begingroup$
Do u know if it is going to be Mensurable for $epsilon$ equal to $1/2$?
$endgroup$
– Pedro Santos
Mar 18 at 8:29
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
If $q_N to 1$, why must we have $U_epsilon subset U_N$?
answered Mar 17 at 12:39
RRLRRL
53k42573
$endgroup$
$begingroup$
Hm i see ur point , that neighborhood can surpass the interval .
$endgroup$
– Pedro Santos
Mar 17 at 12:40
$begingroup$
Also the enumeration cannot be monotonically increasing.
$endgroup$
– RRL
Mar 17 at 12:42
$begingroup$
Do u know if it is going to be Mensurable for $epsilon$ equal to $1/2$?
$endgroup$
– Pedro Santos
Mar 18 at 8:29
add a comment |
$begingroup$
Hint:
If $q_N to 1$, why must we have $U_epsilon subset U_N$?
answered Mar 17 at 12:39
RRLRRL
53k42573
$endgroup$
$begingroup$
Hm i see ur point , that neighborhood can surpass the interval .
$endgroup$
– Pedro Santos
Mar 17 at 12:40
$begingroup$
Also the enumeration cannot be monotonically increasing.
$endgroup$
– RRL
Mar 17 at 12:42
$begingroup$
Do u know if it is going to be Mensurable for $epsilon$ equal to $1/2$?
$endgroup$
– Pedro Santos
Mar 18 at 8:29
add a comment |
$begingroup$
Hint:
If $q_N to 1$, why must we have $U_epsilon subset U_N$?
answered Mar 17 at 12:39
RRLRRL
53k42573
$endgroup$
Hint:
If $q_N to 1$, why must we have $U_epsilon subset U_N$?
answered Mar 17 at 12:39
RRLRRL
53k42573
answered Mar 17 at 12:39
RRLRRL
53k42573
answered Mar 17 at 12:39
RRLRRL
53k42573
answered Mar 17 at 12:39
RRLRRL
53k42573
53k42573
$begingroup$
Hm i see ur point , that neighborhood can surpass the interval .
$endgroup$
– Pedro Santos
Mar 17 at 12:40
$begingroup$
Also the enumeration cannot be monotonically increasing.
$endgroup$
– RRL
Mar 17 at 12:42
$begingroup$
Do u know if it is going to be Mensurable for $epsilon$ equal to $1/2$?
$endgroup$
– Pedro Santos
Mar 18 at 8:29
add a comment |
$begingroup$
Hm i see ur point , that neighborhood can surpass the interval .
$endgroup$
– Pedro Santos
Mar 17 at 12:40
$begingroup$
Also the enumeration cannot be monotonically increasing.
$endgroup$
– RRL
Mar 17 at 12:42
$begingroup$
Do u know if it is going to be Mensurable for $epsilon$ equal to $1/2$?
$endgroup$
– Pedro Santos
Mar 18 at 8:29
$begingroup$
Hm i see ur point , that neighborhood can surpass the interval .
$endgroup$
– Pedro Santos
Mar 17 at 12:40
$begingroup$
Hm i see ur point , that neighborhood can surpass the interval .
$endgroup$
– Pedro Santos
Mar 17 at 12:40
$begingroup$
Also the enumeration cannot be monotonically increasing.
$endgroup$
– RRL
Mar 17 at 12:42
$begingroup$
Also the enumeration cannot be monotonically increasing.
$endgroup$
– RRL
Mar 17 at 12:42
$begingroup$
Do u know if it is going to be Mensurable for $epsilon$ equal to $1/2$?
$endgroup$
– Pedro Santos
Mar 18 at 8:29
$begingroup$
Do u know if it is going to be Mensurable for $epsilon$ equal to $1/2$?
$endgroup$
– Pedro Santos
Mar 18 at 8:29
add a comment |
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