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Collection of all $TsubseteqmathbbR$ such that $λ^*(T) =λ^∗(T∩A)$ for all $AsubseteqmathbbR$.

Are supersets of measurable sets measurable?Are supersets of non-empty measurable sets measurable?Show that $rho(mathbbQ, x) = 0$ and $rho(mathbbR setminus mathbbQ, x) = 1$ for all $x in mathbbR$All the functions $phi$ such that $phi (lim_pto 0||f||_p)=int_0^1 (phi of)dm$Finding an interval on the set that is not m*-measurable.Consider a measurable function $f:[0,1]rightarrow mathbbR$ such that $f(x)geq ||f||_2$.question on showing convergence of an integralLebesgue Measurable Subset of $[0,1]times[0,1]$Show that a given measure is equal to the Lebesgue measure on Borel subsets on $mathbbR$Lebesgue measurable set with certain properties

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$begingroup$

Hi I am doing a past paper and I am stuck on the following question. Let $theta$ be the collection of all $TsubseteqmathbbR$ such that $λ^*(T) =λ^∗(T∩A)$ for all $AsubseteqmathbbR$ , show that $theta$ can be written as $theta$ = $TsubseteqmathbbR$ : $λ^*(T)leq m$ and find $m$. The first part of the question asks to give an example of an element of $theta$ for which I gave the empty set, but I am really unsure of what to do for this part of the question. I am confident that all the elements of $theta$ are the subsets of $A$ but I don't know how I would use this to answer the question. Any help would be appreciated.

lebesgue-measure

share|cite|improve this question

edited Mar 15 at 18:58

Asaf Karagila♦

307k33438769

asked Mar 15 at 18:35

RJSSRJSS

52

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given any $Tintheta$ we have $λ^*(T)=λ^*(T∩T^c)=0$
  $endgroup$
  – Holo
  Mar 15 at 18:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the hint I appreciate it!
  $endgroup$
  – RJSS
  Mar 15 at 20:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Holo Hi, the only thing I can seem to get from this is a trivial lower bound, $λ^∗(T)geq0$. I am unsure of how to obtain an upper bound for $λ^∗(T)$ from this.
  $endgroup$
  – RJSS
  Mar 15 at 21:17
  

  
  
  
  

add a comment | 

0

$begingroup$

Hi I am doing a past paper and I am stuck on the following question. Let $theta$ be the collection of all $TsubseteqmathbbR$ such that $λ^*(T) =λ^∗(T∩A)$ for all $AsubseteqmathbbR$ , show that $theta$ can be written as $theta$ = $TsubseteqmathbbR$ : $λ^*(T)leq m$ and find $m$. The first part of the question asks to give an example of an element of $theta$ for which I gave the empty set, but I am really unsure of what to do for this part of the question. I am confident that all the elements of $theta$ are the subsets of $A$ but I don't know how I would use this to answer the question. Any help would be appreciated.

lebesgue-measure

share|cite|improve this question

edited Mar 15 at 18:58

Asaf Karagila♦

307k33438769

asked Mar 15 at 18:35

RJSSRJSS

52

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given any $Tintheta$ we have $λ^*(T)=λ^*(T∩T^c)=0$
  $endgroup$
  – Holo
  Mar 15 at 18:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the hint I appreciate it!
  $endgroup$
  – RJSS
  Mar 15 at 20:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Holo Hi, the only thing I can seem to get from this is a trivial lower bound, $λ^∗(T)geq0$. I am unsure of how to obtain an upper bound for $λ^∗(T)$ from this.
  $endgroup$
  – RJSS
  Mar 15 at 21:17
  

  
  
  
  

add a comment | 

$begingroup$

Hi I am doing a past paper and I am stuck on the following question. Let $theta$ be the collection of all $TsubseteqmathbbR$ such that $λ^*(T) =λ^∗(T∩A)$ for all $AsubseteqmathbbR$ , show that $theta$ can be written as $theta$ = $TsubseteqmathbbR$ : $λ^*(T)leq m$ and find $m$. The first part of the question asks to give an example of an element of $theta$ for which I gave the empty set, but I am really unsure of what to do for this part of the question. I am confident that all the elements of $theta$ are the subsets of $A$ but I don't know how I would use this to answer the question. Any help would be appreciated.

lebesgue-measure

share|cite|improve this question

edited Mar 15 at 18:58

Asaf Karagila♦

307k33438769

asked Mar 15 at 18:35

RJSSRJSS

52

$endgroup$

share|cite|improve this question

edited Mar 15 at 18:58

Asaf Karagila♦

307k33438769

asked Mar 15 at 18:35

RJSSRJSS

52

share|cite|improve this question

edited Mar 15 at 18:58

Asaf Karagila♦

307k33438769

asked Mar 15 at 18:35

RJSSRJSS

52

edited Mar 15 at 18:58

Asaf Karagila♦

307k33438769

edited Mar 15 at 18:58

Asaf Karagila♦

307k33438769

asked Mar 15 at 18:35

RJSSRJSS

52

asked Mar 15 at 18:35

RJSSRJSS

52

1 Answer
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1

$begingroup$

It is given that $∀T∈θ:∀A⊆mathbb R~(λ^*(T)=λ^*(T∩A))$ in particular $λ^*(T)=λ^*(T∩T^c)$ but $λ^*(T∩T^c)=0$ so $λ^*(T)=0$.

So with that we get that $T∈θimplies T∈B⊆mathbb Rmid λ^*(B)=0$

Now given $T ∈B⊆mathbb Rmid λ^*(A)=0$, we know that $T'⊆T$ implies $λ^*(T')=0$, in addition $T∩A⊆T$, so $λ^*(T)=λ^*(T∩A)$ for all $A⊆mathbb R$, thus $T∈θ$.

So $θ=B⊆mathbb Rmid λ^*(B)=0=B⊆mathbb Rmid λ^*(B)≤0$

share|cite|improve this answer

answered Mar 15 at 21:31

HoloHolo

6,08421131

$endgroup$

add a comment | 

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1

$begingroup$

It is given that $∀T∈θ:∀A⊆mathbb R~(λ^*(T)=λ^*(T∩A))$ in particular $λ^*(T)=λ^*(T∩T^c)$ but $λ^*(T∩T^c)=0$ so $λ^*(T)=0$.

So with that we get that $T∈θimplies T∈B⊆mathbb Rmid λ^*(B)=0$

Now given $T ∈B⊆mathbb Rmid λ^*(A)=0$, we know that $T'⊆T$ implies $λ^*(T')=0$, in addition $T∩A⊆T$, so $λ^*(T)=λ^*(T∩A)$ for all $A⊆mathbb R$, thus $T∈θ$.

So $θ=B⊆mathbb Rmid λ^*(B)=0=B⊆mathbb Rmid λ^*(B)≤0$

share|cite|improve this answer

answered Mar 15 at 21:31

HoloHolo

6,08421131

$endgroup$

add a comment | 

1

$begingroup$

It is given that $∀T∈θ:∀A⊆mathbb R~(λ^*(T)=λ^*(T∩A))$ in particular $λ^*(T)=λ^*(T∩T^c)$ but $λ^*(T∩T^c)=0$ so $λ^*(T)=0$.

So with that we get that $T∈θimplies T∈B⊆mathbb Rmid λ^*(B)=0$

Now given $T ∈B⊆mathbb Rmid λ^*(A)=0$, we know that $T'⊆T$ implies $λ^*(T')=0$, in addition $T∩A⊆T$, so $λ^*(T)=λ^*(T∩A)$ for all $A⊆mathbb R$, thus $T∈θ$.

So $θ=B⊆mathbb Rmid λ^*(B)=0=B⊆mathbb Rmid λ^*(B)≤0$

share|cite|improve this answer

answered Mar 15 at 21:31

HoloHolo

6,08421131

$endgroup$

add a comment | 

$begingroup$

It is given that $∀T∈θ:∀A⊆mathbb R~(λ^*(T)=λ^*(T∩A))$ in particular $λ^*(T)=λ^*(T∩T^c)$ but $λ^*(T∩T^c)=0$ so $λ^*(T)=0$.

So with that we get that $T∈θimplies T∈B⊆mathbb Rmid λ^*(B)=0$

Now given $T ∈B⊆mathbb Rmid λ^*(A)=0$, we know that $T'⊆T$ implies $λ^*(T')=0$, in addition $T∩A⊆T$, so $λ^*(T)=λ^*(T∩A)$ for all $A⊆mathbb R$, thus $T∈θ$.

So $θ=B⊆mathbb Rmid λ^*(B)=0=B⊆mathbb Rmid λ^*(B)≤0$

share|cite|improve this answer

answered Mar 15 at 21:31

HoloHolo

6,08421131

$endgroup$

share|cite|improve this answer

answered Mar 15 at 21:31

HoloHolo

6,08421131

answered Mar 15 at 21:31

HoloHolo

6,08421131

answered Mar 15 at 21:31

HoloHolo

6,08421131