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Middle Fifths Cantor Set is Borel and Has Measure =?
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I've been working on the following problem; I think I have the answer, I would just like to confirm that there are no gaps in my logic.
Consider the "middle fifths" Cantor set $$mathscrC=leftlbracesum_j=1^inftyfraca_j5^j~:~a_j=0,1,3,mathrmor~4~mathrmfor~each~jrightrbrace.$$ Prove that $mathscrC$ is a Borel set. What is the Lebesgue measure of $mathscrC$?
My Solution: We consider the construction of the above Cantor set as follows: let $C_0=[0,1]$, let $C_1=[0,2/5]cup[3/5,1]$, and in general, let $C_n$ be the set remaining after removing an open interval of length $1/5$ from the middle of each connected component of $C_n-1$. Then we put $mathscrC=bigcap_n=0^inftyC_n$.
Let $mathscrB$ be the Borel $sigma$-algebra on $mathbbR$. Then, by definition, $mathscrB$ contains all the open subsets of $mathbbR$. Since $sigma$-algebras are closed under set complements, it follows that $mathscrB$ also contains all the closed subsets of $mathbbR$. Hence, $C_ninmathscrB$ for each $n$. Moreover, since $sigma$-algebras are closed under countable intersections and $C_ninmathscrB$ for each $n$, it follows that $mathscrC=bigcap_n=0^inftyC_n$ is Borel.
We claim that $m(mathscrC)=0$, where $m$ is the Lebesgue measure on $mathbbR$. Indeed, each set $C_n$ is the disjoint union of precisely $2^n$ intervals, each of length $1/5^n$. By the finite additivity of the Lebesgue measure, we have $m(C_n)=2^n(1/5^n)=(2/5)^n$. Also, by the monotonicity of the Lebesgue measure, we have $$m(mathscrC)leq m(C_n)=(2/5)^n.$$ Since the inequality above holds for all $n$, we conclude that $m(mathscrC)=0$.
Does all of this look okay? I wanted to make sure, since I didn't end up using the original description of $mathscrC$ that was given in the problem statement. Thanks in advance for any help!
Edit: After reading up on Fat Cantor Sets, I think my work above is incorrect. Should this Cantor set have positive measure?
real-analysis measure-theory proof-verification lebesgue-measure cantor-set
edited 32 mins ago
Martin Sleziak
43.9k6113264
asked 1 hour ago
Sir_Math_Cat
907618
add a comment |Â
up vote
1
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I've been working on the following problem; I think I have the answer, I would just like to confirm that there are no gaps in my logic.
Consider the "middle fifths" Cantor set $$mathscrC=leftlbracesum_j=1^inftyfraca_j5^j~:~a_j=0,1,3,mathrmor~4~mathrmfor~each~jrightrbrace.$$ Prove that $mathscrC$ is a Borel set. What is the Lebesgue measure of $mathscrC$?
My Solution: We consider the construction of the above Cantor set as follows: let $C_0=[0,1]$, let $C_1=[0,2/5]cup[3/5,1]$, and in general, let $C_n$ be the set remaining after removing an open interval of length $1/5$ from the middle of each connected component of $C_n-1$. Then we put $mathscrC=bigcap_n=0^inftyC_n$.
Let $mathscrB$ be the Borel $sigma$-algebra on $mathbbR$. Then, by definition, $mathscrB$ contains all the open subsets of $mathbbR$. Since $sigma$-algebras are closed under set complements, it follows that $mathscrB$ also contains all the closed subsets of $mathbbR$. Hence, $C_ninmathscrB$ for each $n$. Moreover, since $sigma$-algebras are closed under countable intersections and $C_ninmathscrB$ for each $n$, it follows that $mathscrC=bigcap_n=0^inftyC_n$ is Borel.
We claim that $m(mathscrC)=0$, where $m$ is the Lebesgue measure on $mathbbR$. Indeed, each set $C_n$ is the disjoint union of precisely $2^n$ intervals, each of length $1/5^n$. By the finite additivity of the Lebesgue measure, we have $m(C_n)=2^n(1/5^n)=(2/5)^n$. Also, by the monotonicity of the Lebesgue measure, we have $$m(mathscrC)leq m(C_n)=(2/5)^n.$$ Since the inequality above holds for all $n$, we conclude that $m(mathscrC)=0$.
Does all of this look okay? I wanted to make sure, since I didn't end up using the original description of $mathscrC$ that was given in the problem statement. Thanks in advance for any help!
Edit: After reading up on Fat Cantor Sets, I think my work above is incorrect. Should this Cantor set have positive measure?
real-analysis measure-theory proof-verification lebesgue-measure cantor-set
edited 32 mins ago
Martin Sleziak
43.9k6113264
asked 1 hour ago
Sir_Math_Cat
907618
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I've been working on the following problem; I think I have the answer, I would just like to confirm that there are no gaps in my logic.
Consider the "middle fifths" Cantor set $$mathscrC=leftlbracesum_j=1^inftyfraca_j5^j~:~a_j=0,1,3,mathrmor~4~mathrmfor~each~jrightrbrace.$$ Prove that $mathscrC$ is a Borel set. What is the Lebesgue measure of $mathscrC$?
My Solution: We consider the construction of the above Cantor set as follows: let $C_0=[0,1]$, let $C_1=[0,2/5]cup[3/5,1]$, and in general, let $C_n$ be the set remaining after removing an open interval of length $1/5$ from the middle of each connected component of $C_n-1$. Then we put $mathscrC=bigcap_n=0^inftyC_n$.
Let $mathscrB$ be the Borel $sigma$-algebra on $mathbbR$. Then, by definition, $mathscrB$ contains all the open subsets of $mathbbR$. Since $sigma$-algebras are closed under set complements, it follows that $mathscrB$ also contains all the closed subsets of $mathbbR$. Hence, $C_ninmathscrB$ for each $n$. Moreover, since $sigma$-algebras are closed under countable intersections and $C_ninmathscrB$ for each $n$, it follows that $mathscrC=bigcap_n=0^inftyC_n$ is Borel.
We claim that $m(mathscrC)=0$, where $m$ is the Lebesgue measure on $mathbbR$. Indeed, each set $C_n$ is the disjoint union of precisely $2^n$ intervals, each of length $1/5^n$. By the finite additivity of the Lebesgue measure, we have $m(C_n)=2^n(1/5^n)=(2/5)^n$. Also, by the monotonicity of the Lebesgue measure, we have $$m(mathscrC)leq m(C_n)=(2/5)^n.$$ Since the inequality above holds for all $n$, we conclude that $m(mathscrC)=0$.
Does all of this look okay? I wanted to make sure, since I didn't end up using the original description of $mathscrC$ that was given in the problem statement. Thanks in advance for any help!
Edit: After reading up on Fat Cantor Sets, I think my work above is incorrect. Should this Cantor set have positive measure?
real-analysis measure-theory proof-verification lebesgue-measure cantor-set
edited 32 mins ago
Martin Sleziak
43.9k6113264
asked 1 hour ago
Sir_Math_Cat
907618
I've been working on the following problem; I think I have the answer, I would just like to confirm that there are no gaps in my logic.
Consider the "middle fifths" Cantor set $$mathscrC=leftlbracesum_j=1^inftyfraca_j5^j~:~a_j=0,1,3,mathrmor~4~mathrmfor~each~jrightrbrace.$$ Prove that $mathscrC$ is a Borel set. What is the Lebesgue measure of $mathscrC$?
My Solution: We consider the construction of the above Cantor set as follows: let $C_0=[0,1]$, let $C_1=[0,2/5]cup[3/5,1]$, and in general, let $C_n$ be the set remaining after removing an open interval of length $1/5$ from the middle of each connected component of $C_n-1$. Then we put $mathscrC=bigcap_n=0^inftyC_n$.
Let $mathscrB$ be the Borel $sigma$-algebra on $mathbbR$. Then, by definition, $mathscrB$ contains all the open subsets of $mathbbR$. Since $sigma$-algebras are closed under set complements, it follows that $mathscrB$ also contains all the closed subsets of $mathbbR$. Hence, $C_ninmathscrB$ for each $n$. Moreover, since $sigma$-algebras are closed under countable intersections and $C_ninmathscrB$ for each $n$, it follows that $mathscrC=bigcap_n=0^inftyC_n$ is Borel.
We claim that $m(mathscrC)=0$, where $m$ is the Lebesgue measure on $mathbbR$. Indeed, each set $C_n$ is the disjoint union of precisely $2^n$ intervals, each of length $1/5^n$. By the finite additivity of the Lebesgue measure, we have $m(C_n)=2^n(1/5^n)=(2/5)^n$. Also, by the monotonicity of the Lebesgue measure, we have $$m(mathscrC)leq m(C_n)=(2/5)^n.$$ Since the inequality above holds for all $n$, we conclude that $m(mathscrC)=0$.
Does all of this look okay? I wanted to make sure, since I didn't end up using the original description of $mathscrC$ that was given in the problem statement. Thanks in advance for any help!
Edit: After reading up on Fat Cantor Sets, I think my work above is incorrect. Should this Cantor set have positive measure?
real-analysis measure-theory proof-verification lebesgue-measure cantor-set
real-analysis measure-theory proof-verification lebesgue-measure cantor-set
edited 32 mins ago
Martin Sleziak
43.9k6113264
asked 1 hour ago
Sir_Math_Cat
907618
edited 32 mins ago
Martin Sleziak
43.9k6113264
asked 1 hour ago
Sir_Math_Cat
907618
edited 32 mins ago
Martin Sleziak
43.9k6113264
edited 32 mins ago
Martin Sleziak
43.9k6113264
edited 32 mins ago
Martin Sleziak
43.9k6113264
43.9k6113264
asked 1 hour ago
Sir_Math_Cat
907618
asked 1 hour ago
Sir_Math_Cat
907618
asked 1 hour ago
Sir_Math_Cat
907618
907618
add a comment |Â
add a comment |Â
2 Answers
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3
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$newcommandCmathscr C$
"I didn't end up using the original description of $mathscr C$ that was given in the problem statement": You used that in showing that $C$ is the same as the $C$ you get by that middle-fifths construction.
Or you should have! In fact the iterative construction you sketch, removing the middle fifth of each component at each stage, is wrong.
You start by removing the middle fifth of $[0,1]$, leaving $[0,2/5]cup[3/5,1]$. That's great - that removes the numbers $sum a_j/5^j$ with $a_1=2$. Now at the next stage you want to remove the numbers with $a_2=2$. To do that you need to remove the middle fifth from each of the four intervals $[0,1/5],[1/5,2/5],[3/5,4/5],[4/5,1]$, which is not the same as removing the middle fifth of each component.
(Yes, the measure is $0$. The simplest way to see that is to note that at each stage the measure gets multiplied by $4/5$. A "fat Cantor set" requires a different ratio at each step.)
answered 28 mins ago
David C. Ullrich
56.6k43788
add a comment |Â
up vote
1
down vote
The computation of measure is not correct. It turns out that it is equal to $frac23$, because$$mleft(mathscrC^complementright)=frac12sum_n=1^inftyleft(frac25right)^n=fracfrac151-frac25=frac13.$$Note that, at each step, you remove $2^n-1$ intervals whose length is $frac15^n$.
answered 1 hour ago
José Carlos Santos
131k17106192
That makes sense. Thank you for your help! Did my argument to show that $mathscrC$ is Borel look okay?
– Sir_Math_Cat
1 hour ago
@Sir_Math_Cat It's just fine.
– José Carlos Santos
1 hour ago
1
Quick question, how does the law of large numbers not imply the measure is 0? Unless I’m misreading something, we are basically looking at the probability of never rolling a 2 if we roll a 5 sided die forever.
– guy
1 hour ago
Okay, thank you very much!
– Sir_Math_Cat
1 hour ago
1
Or simpler: At each stage the measure gets multiplied by $4/5$.
– David C. Ullrich
45 mins ago
 |Â
show 7 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
$newcommandCmathscr C$
"I didn't end up using the original description of $mathscr C$ that was given in the problem statement": You used that in showing that $C$ is the same as the $C$ you get by that middle-fifths construction.
Or you should have! In fact the iterative construction you sketch, removing the middle fifth of each component at each stage, is wrong.
You start by removing the middle fifth of $[0,1]$, leaving $[0,2/5]cup[3/5,1]$. That's great - that removes the numbers $sum a_j/5^j$ with $a_1=2$. Now at the next stage you want to remove the numbers with $a_2=2$. To do that you need to remove the middle fifth from each of the four intervals $[0,1/5],[1/5,2/5],[3/5,4/5],[4/5,1]$, which is not the same as removing the middle fifth of each component.
(Yes, the measure is $0$. The simplest way to see that is to note that at each stage the measure gets multiplied by $4/5$. A "fat Cantor set" requires a different ratio at each step.)
answered 28 mins ago
David C. Ullrich
56.6k43788
add a comment |Â
up vote
3
down vote
$newcommandCmathscr C$
"I didn't end up using the original description of $mathscr C$ that was given in the problem statement": You used that in showing that $C$ is the same as the $C$ you get by that middle-fifths construction.
Or you should have! In fact the iterative construction you sketch, removing the middle fifth of each component at each stage, is wrong.
You start by removing the middle fifth of $[0,1]$, leaving $[0,2/5]cup[3/5,1]$. That's great - that removes the numbers $sum a_j/5^j$ with $a_1=2$. Now at the next stage you want to remove the numbers with $a_2=2$. To do that you need to remove the middle fifth from each of the four intervals $[0,1/5],[1/5,2/5],[3/5,4/5],[4/5,1]$, which is not the same as removing the middle fifth of each component.
(Yes, the measure is $0$. The simplest way to see that is to note that at each stage the measure gets multiplied by $4/5$. A "fat Cantor set" requires a different ratio at each step.)
answered 28 mins ago
David C. Ullrich
56.6k43788
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$newcommandCmathscr C$
"I didn't end up using the original description of $mathscr C$ that was given in the problem statement": You used that in showing that $C$ is the same as the $C$ you get by that middle-fifths construction.
Or you should have! In fact the iterative construction you sketch, removing the middle fifth of each component at each stage, is wrong.
You start by removing the middle fifth of $[0,1]$, leaving $[0,2/5]cup[3/5,1]$. That's great - that removes the numbers $sum a_j/5^j$ with $a_1=2$. Now at the next stage you want to remove the numbers with $a_2=2$. To do that you need to remove the middle fifth from each of the four intervals $[0,1/5],[1/5,2/5],[3/5,4/5],[4/5,1]$, which is not the same as removing the middle fifth of each component.
(Yes, the measure is $0$. The simplest way to see that is to note that at each stage the measure gets multiplied by $4/5$. A "fat Cantor set" requires a different ratio at each step.)
answered 28 mins ago
David C. Ullrich
56.6k43788
$newcommandCmathscr C$
"I didn't end up using the original description of $mathscr C$ that was given in the problem statement": You used that in showing that $C$ is the same as the $C$ you get by that middle-fifths construction.
Or you should have! In fact the iterative construction you sketch, removing the middle fifth of each component at each stage, is wrong.
You start by removing the middle fifth of $[0,1]$, leaving $[0,2/5]cup[3/5,1]$. That's great - that removes the numbers $sum a_j/5^j$ with $a_1=2$. Now at the next stage you want to remove the numbers with $a_2=2$. To do that you need to remove the middle fifth from each of the four intervals $[0,1/5],[1/5,2/5],[3/5,4/5],[4/5,1]$, which is not the same as removing the middle fifth of each component.
(Yes, the measure is $0$. The simplest way to see that is to note that at each stage the measure gets multiplied by $4/5$. A "fat Cantor set" requires a different ratio at each step.)
answered 28 mins ago
David C. Ullrich
56.6k43788
edited 11 mins ago
answered 28 mins ago
David C. Ullrich
56.6k43788
answered 28 mins ago
David C. Ullrich
56.6k43788
answered 28 mins ago
David C. Ullrich
56.6k43788
56.6k43788
add a comment |Â
add a comment |Â
up vote
1
down vote
The computation of measure is not correct. It turns out that it is equal to $frac23$, because$$mleft(mathscrC^complementright)=frac12sum_n=1^inftyleft(frac25right)^n=fracfrac151-frac25=frac13.$$Note that, at each step, you remove $2^n-1$ intervals whose length is $frac15^n$.
answered 1 hour ago
José Carlos Santos
131k17106192
That makes sense. Thank you for your help! Did my argument to show that $mathscrC$ is Borel look okay?
– Sir_Math_Cat
1 hour ago
@Sir_Math_Cat It's just fine.
– José Carlos Santos
1 hour ago
1
Quick question, how does the law of large numbers not imply the measure is 0? Unless I’m misreading something, we are basically looking at the probability of never rolling a 2 if we roll a 5 sided die forever.
– guy
1 hour ago
Okay, thank you very much!
– Sir_Math_Cat
1 hour ago
1
Or simpler: At each stage the measure gets multiplied by $4/5$.
– David C. Ullrich
45 mins ago
 |Â
show 7 more comments
up vote
1
down vote
The computation of measure is not correct. It turns out that it is equal to $frac23$, because$$mleft(mathscrC^complementright)=frac12sum_n=1^inftyleft(frac25right)^n=fracfrac151-frac25=frac13.$$Note that, at each step, you remove $2^n-1$ intervals whose length is $frac15^n$.
answered 1 hour ago
José Carlos Santos
131k17106192
That makes sense. Thank you for your help! Did my argument to show that $mathscrC$ is Borel look okay?
– Sir_Math_Cat
1 hour ago
@Sir_Math_Cat It's just fine.
– José Carlos Santos
1 hour ago
1
Quick question, how does the law of large numbers not imply the measure is 0? Unless I’m misreading something, we are basically looking at the probability of never rolling a 2 if we roll a 5 sided die forever.
– guy
1 hour ago
Okay, thank you very much!
– Sir_Math_Cat
1 hour ago
1
Or simpler: At each stage the measure gets multiplied by $4/5$.
– David C. Ullrich
45 mins ago
 |Â
show 7 more comments
up vote
1
down vote
up vote
1
down vote
The computation of measure is not correct. It turns out that it is equal to $frac23$, because$$mleft(mathscrC^complementright)=frac12sum_n=1^inftyleft(frac25right)^n=fracfrac151-frac25=frac13.$$Note that, at each step, you remove $2^n-1$ intervals whose length is $frac15^n$.
answered 1 hour ago
José Carlos Santos
131k17106192
The computation of measure is not correct. It turns out that it is equal to $frac23$, because$$mleft(mathscrC^complementright)=frac12sum_n=1^inftyleft(frac25right)^n=fracfrac151-frac25=frac13.$$Note that, at each step, you remove $2^n-1$ intervals whose length is $frac15^n$.
answered 1 hour ago
José Carlos Santos
131k17106192
answered 1 hour ago
José Carlos Santos
131k17106192
answered 1 hour ago
José Carlos Santos
131k17106192
answered 1 hour ago
José Carlos Santos
131k17106192
131k17106192
That makes sense. Thank you for your help! Did my argument to show that $mathscrC$ is Borel look okay?
– Sir_Math_Cat
1 hour ago
@Sir_Math_Cat It's just fine.
– José Carlos Santos
1 hour ago
1
Quick question, how does the law of large numbers not imply the measure is 0? Unless I’m misreading something, we are basically looking at the probability of never rolling a 2 if we roll a 5 sided die forever.
– guy
1 hour ago
Okay, thank you very much!
– Sir_Math_Cat
1 hour ago
1
Or simpler: At each stage the measure gets multiplied by $4/5$.
– David C. Ullrich
45 mins ago
 |Â
show 7 more comments
That makes sense. Thank you for your help! Did my argument to show that $mathscrC$ is Borel look okay?
– Sir_Math_Cat
1 hour ago
@Sir_Math_Cat It's just fine.
– José Carlos Santos
1 hour ago
1
Quick question, how does the law of large numbers not imply the measure is 0? Unless I’m misreading something, we are basically looking at the probability of never rolling a 2 if we roll a 5 sided die forever.
– guy
1 hour ago
Okay, thank you very much!
– Sir_Math_Cat
1 hour ago
1
Or simpler: At each stage the measure gets multiplied by $4/5$.
– David C. Ullrich
45 mins ago
That makes sense. Thank you for your help! Did my argument to show that $mathscrC$ is Borel look okay?
– Sir_Math_Cat
1 hour ago
That makes sense. Thank you for your help! Did my argument to show that $mathscrC$ is Borel look okay?
– Sir_Math_Cat
1 hour ago
@Sir_Math_Cat It's just fine.
– José Carlos Santos
1 hour ago
@Sir_Math_Cat It's just fine.
– José Carlos Santos
1 hour ago
1
1
Quick question, how does the law of large numbers not imply the measure is 0? Unless I’m misreading something, we are basically looking at the probability of never rolling a 2 if we roll a 5 sided die forever.
– guy
1 hour ago
Quick question, how does the law of large numbers not imply the measure is 0? Unless I’m misreading something, we are basically looking at the probability of never rolling a 2 if we roll a 5 sided die forever.
– guy
1 hour ago
Okay, thank you very much!
– Sir_Math_Cat
1 hour ago
Okay, thank you very much!
– Sir_Math_Cat
1 hour ago
1
1
Or simpler: At each stage the measure gets multiplied by $4/5$.
– David C. Ullrich
45 mins ago
Or simpler: At each stage the measure gets multiplied by $4/5$.
– David C. Ullrich
45 mins ago
 |Â
show 7 more comments
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