Is a set in between two sets of equal measure measurable?

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The Lebesgue sigma algebra is complete with respect to Lebesgue measure, which means that if $A$ is a Lebesgue measurable set with Lebesgue measure $0$ and $B$ is a subset of $A$, then $B$ is Lebesgue measurable as well. But I'd like to know if something stronger is true.

Suppose that $A$ is a subset of $B$ which is a subset of $C$, where $A$ and $C$ are Lebesgue measurable sets and the Lebesgue measure of $A$ is equal to the Lebesgue measure of $C$. Then my question is, does $B$ have to be Lebesgue measurable as well?

measure-theory lebesgue-measure

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asked 15 mins ago

Keshav Srinivasan

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The Lebesgue sigma algebra is complete with respect to Lebesgue measure, which means that if $A$ is a Lebesgue measurable set with Lebesgue measure $0$ and $B$ is a subset of $A$, then $B$ is Lebesgue measurable as well. But I'd like to know if something stronger is true.

Suppose that $A$ is a subset of $B$ which is a subset of $C$, where $A$ and $C$ are Lebesgue measurable sets and the Lebesgue measure of $A$ is equal to the Lebesgue measure of $C$. Then my question is, does $B$ have to be Lebesgue measurable as well?

measure-theory lebesgue-measure

share|cite|improve this question

asked 15 mins ago

Keshav Srinivasan

2,28611340

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

The Lebesgue sigma algebra is complete with respect to Lebesgue measure, which means that if $A$ is a Lebesgue measurable set with Lebesgue measure $0$ and $B$ is a subset of $A$, then $B$ is Lebesgue measurable as well. But I'd like to know if something stronger is true.

Suppose that $A$ is a subset of $B$ which is a subset of $C$, where $A$ and $C$ are Lebesgue measurable sets and the Lebesgue measure of $A$ is equal to the Lebesgue measure of $C$. Then my question is, does $B$ have to be Lebesgue measurable as well?

measure-theory lebesgue-measure

share|cite|improve this question

asked 15 mins ago

Keshav Srinivasan

2,28611340

The Lebesgue sigma algebra is complete with respect to Lebesgue measure, which means that if $A$ is a Lebesgue measurable set with Lebesgue measure $0$ and $B$ is a subset of $A$, then $B$ is Lebesgue measurable as well. But I'd like to know if something stronger is true.

Suppose that $A$ is a subset of $B$ which is a subset of $C$, where $A$ and $C$ are Lebesgue measurable sets and the Lebesgue measure of $A$ is equal to the Lebesgue measure of $C$. Then my question is, does $B$ have to be Lebesgue measurable as well?

measure-theory lebesgue-measure

measure-theory lebesgue-measure

share|cite|improve this question

asked 15 mins ago

Keshav Srinivasan

2,28611340

share|cite|improve this question

asked 15 mins ago

Keshav Srinivasan

2,28611340

share|cite|improve this question

share|cite|improve this question

asked 15 mins ago

Keshav Srinivasan

2,28611340

asked 15 mins ago

Keshav Srinivasan

2,28611340

asked 15 mins ago

Keshav Srinivasan

2,28611340

2,28611340

add a comment | 

add a comment | 

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If $A$ and $C$ are each Lebesgue measurable, then so is $Csetminus A$ and so by finite additivity it has measure zero.

But then $Bsetminus A$ also has measure zero and hence is measurable, so $$B=Acup(Bsetminus A)$$ is the union of two measurable sets, hence is measurable.

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answered 12 mins ago

Noah Schweber

116k10143274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, that was more trivial than I thought. What about this: suppose that for any $epsilon>0$, there exist Lebesgue measurable sets $A$ and $C$ such that $A$ is a subset of $B$ which is a subset of $C$ and the Lebesgue measure of $C$ minus the Lebesgue measure of $A$ is less than $epsilon$. Then does $B$ have to be Lebesgue measurable?
  – Keshav Srinivasan
  just now
  

  
  
  
  

add a comment | 

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1 Answer
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1 Answer
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active

oldest

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active

oldest

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active

oldest

votes

up vote
5
down vote

If $A$ and $C$ are each Lebesgue measurable, then so is $Csetminus A$ and so by finite additivity it has measure zero.

But then $Bsetminus A$ also has measure zero and hence is measurable, so $$B=Acup(Bsetminus A)$$ is the union of two measurable sets, hence is measurable.

share|cite|improve this answer

answered 12 mins ago

Noah Schweber

116k10143274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, that was more trivial than I thought. What about this: suppose that for any $epsilon>0$, there exist Lebesgue measurable sets $A$ and $C$ such that $A$ is a subset of $B$ which is a subset of $C$ and the Lebesgue measure of $C$ minus the Lebesgue measure of $A$ is less than $epsilon$. Then does $B$ have to be Lebesgue measurable?
  – Keshav Srinivasan
  just now
  

  
  
  
  

add a comment | 

up vote
5
down vote

If $A$ and $C$ are each Lebesgue measurable, then so is $Csetminus A$ and so by finite additivity it has measure zero.

But then $Bsetminus A$ also has measure zero and hence is measurable, so $$B=Acup(Bsetminus A)$$ is the union of two measurable sets, hence is measurable.

share|cite|improve this answer

answered 12 mins ago

Noah Schweber

116k10143274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, that was more trivial than I thought. What about this: suppose that for any $epsilon>0$, there exist Lebesgue measurable sets $A$ and $C$ such that $A$ is a subset of $B$ which is a subset of $C$ and the Lebesgue measure of $C$ minus the Lebesgue measure of $A$ is less than $epsilon$. Then does $B$ have to be Lebesgue measurable?
  – Keshav Srinivasan
  just now
  

  
  
  
  

add a comment | 

up vote
5
down vote

up vote
5
down vote

If $A$ and $C$ are each Lebesgue measurable, then so is $Csetminus A$ and so by finite additivity it has measure zero.

But then $Bsetminus A$ also has measure zero and hence is measurable, so $$B=Acup(Bsetminus A)$$ is the union of two measurable sets, hence is measurable.

share|cite|improve this answer

answered 12 mins ago

Noah Schweber

116k10143274

If $A$ and $C$ are each Lebesgue measurable, then so is $Csetminus A$ and so by finite additivity it has measure zero.

But then $Bsetminus A$ also has measure zero and hence is measurable, so $$B=Acup(Bsetminus A)$$ is the union of two measurable sets, hence is measurable.

share|cite|improve this answer

answered 12 mins ago

Noah Schweber

116k10143274

share|cite|improve this answer

share|cite|improve this answer

answered 12 mins ago

Noah Schweber

116k10143274

answered 12 mins ago

Noah Schweber

116k10143274

answered 12 mins ago

Noah Schweber

116k10143274

116k10143274

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, that was more trivial than I thought. What about this: suppose that for any $epsilon>0$, there exist Lebesgue measurable sets $A$ and $C$ such that $A$ is a subset of $B$ which is a subset of $C$ and the Lebesgue measure of $C$ minus the Lebesgue measure of $A$ is less than $epsilon$. Then does $B$ have to be Lebesgue measurable?
  – Keshav Srinivasan
  just now
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, that was more trivial than I thought. What about this: suppose that for any $epsilon>0$, there exist Lebesgue measurable sets $A$ and $C$ such that $A$ is a subset of $B$ which is a subset of $C$ and the Lebesgue measure of $C$ minus the Lebesgue measure of $A$ is less than $epsilon$. Then does $B$ have to be Lebesgue measurable?
  – Keshav Srinivasan
  just now
  

  
  
  
  

Thanks, that was more trivial than I thought. What about this: suppose that for any $epsilon>0$, there exist Lebesgue measurable sets $A$ and $C$ such that $A$ is a subset of $B$ which is a subset of $C$ and the Lebesgue measure of $C$ minus the Lebesgue measure of $A$ is less than $epsilon$. Then does $B$ have to be Lebesgue measurable?
– Keshav Srinivasan
just now

Thanks, that was more trivial than I thought. What about this: suppose that for any $epsilon>0$, there exist Lebesgue measurable sets $A$ and $C$ such that $A$ is a subset of $B$ which is a subset of $C$ and the Lebesgue measure of $C$ minus the Lebesgue measure of $A$ is less than $epsilon$. Then does $B$ have to be Lebesgue measurable?
– Keshav Srinivasan
just now

add a comment | 

 

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