Dtjytyk

How to make readers know that my work has used a hidden constraint?

What to do when during a meeting client people start to fight (even physically) with each others?

My story is written in English, but is set in my home country. What language should I use for the dialogue?

Welcoming 2019 Pi day: How to draw the letter π?

How could a female member of a species produce eggs unto death?

Is having access to past exams cheating and, if yes, could it be proven just by a good grade?

Can "semicircle" be used to refer to a part-circle that is not a exact half-circle?

"However" used in a conditional clause?

What does it mean when multiple 々 marks follow a 、?

What has been your most complicated TikZ drawing?

What wound would be of little consequence to a biped but terrible for a quadruped?

Do Bugbears' arms literally get longer when it's their turn?

Running a subshell from the middle of the current command

How can I discourage/prevent PCs from using door choke-points?

When two POV characters meet

Rejected in 4th interview round citing insufficient years of experience

Best mythical creature to use as livestock?

What Happens when Passenger Refuses to Fly Boeing 737 Max?

What is the blue range indicating on this manifold pressure gauge?

Why would a jet engine that runs at temps excess of 2000°C burn when it crashes?

Unreachable code, but reachable with exception

Good allowance savings plan?

Counter-example to the existence of left Bousfield localization of combinatorial model category

"One can do his homework in the library"

Let $f: mathbb{R} to mathbb{R}$ be continuous and let $g: mathbb{R} to mathbb{R}$ Lebesgue measurable. Is $f circ g$ Borel measurable?

Lebesgue and Borel MeasurableBorel measurable function that preserves Lebesgue measureProve that there exists a Borel measurable function $h: mathbb R to mathbb R$ such that $g= hcirc f$.Show that inverse image of a Lebesgue measurable function is Lebesgue-measurableComposition of measurable & continuous functions, is it measurable?Composition of 2 Lebesgue measurable functions is not lebesgue measurable: Are these two functions Borel Measurable?Borel Measurable Function but not Lebesgue MeasurableProve that there is exist Lebesgue measurable and non Borel measurable sets.Not all sets in $mathbb{R}$ are Lebesgue measurable setsIs every Borel subset of a measurable set measurable?

2

$begingroup$

Let $f: mathbb{R} to mathbb{R}$ be continuous and $g: mathbb{R} to mathbb{R}$ Lebesgue measurable. Is $f circ g$ Borel measurable?

I think that the answer to this question is no. I know that the hypothesis implies that $f circ g$ is Lebesgue measurable, but I am sure that there is a counter-example to show that $f circ g$ is not Borel measurable, since the Borel sigma algebra is a proper subset of the Lebesgue sigma algebra.

Note: In my textbook, a function is measurable with respect to a sigma algebra $mathcal{E}$ if and only if $[f > alpha] = {x : f(x) > alpha} in mathcal{E}$ for all $alpha in mathbb{R}$.

My current idea is to find a Lebesgue measurable set $D$ that is not Borel measurable. Then define $g(x) = chi_{_{D}}$, which is Lebesgue measurable, as $D$ is Lebesgue measurable. Then $f(x) = x$ is continuous and $f circ g = g$, but $g$ is not Borel measurable.

measure-theory

share|cite|improve this question

edited Mar 10 at 3:05

johnny133253

asked Mar 10 at 2:02

johnny133253johnny133253

463111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does $X=mathbb{R}$?
  $endgroup$
  – confused_wallet
  Mar 10 at 3:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that was a typo. Fixed.
  $endgroup$
  – johnny133253
  Mar 10 at 3:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I think your idea is a good one!
  $endgroup$
  – confused_wallet
  Mar 10 at 3:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your current idea is correct.
  $endgroup$
  – Ramiro
  Mar 10 at 15:31
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $f: mathbb{R} to mathbb{R}$ be continuous and $g: mathbb{R} to mathbb{R}$ Lebesgue measurable. Is $f circ g$ Borel measurable?

I think that the answer to this question is no. I know that the hypothesis implies that $f circ g$ is Lebesgue measurable, but I am sure that there is a counter-example to show that $f circ g$ is not Borel measurable, since the Borel sigma algebra is a proper subset of the Lebesgue sigma algebra.

Note: In my textbook, a function is measurable with respect to a sigma algebra $mathcal{E}$ if and only if $[f > alpha] = {x : f(x) > alpha} in mathcal{E}$ for all $alpha in mathbb{R}$.

My current idea is to find a Lebesgue measurable set $D$ that is not Borel measurable. Then define $g(x) = chi_{_{D}}$, which is Lebesgue measurable, as $D$ is Lebesgue measurable. Then $f(x) = x$ is continuous and $f circ g = g$, but $g$ is not Borel measurable.

measure-theory

share|cite|improve this question

edited Mar 10 at 3:05

johnny133253

asked Mar 10 at 2:02

johnny133253johnny133253

463111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does $X=mathbb{R}$?
  $endgroup$
  – confused_wallet
  Mar 10 at 3:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that was a typo. Fixed.
  $endgroup$
  – johnny133253
  Mar 10 at 3:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I think your idea is a good one!
  $endgroup$
  – confused_wallet
  Mar 10 at 3:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your current idea is correct.
  $endgroup$
  – Ramiro
  Mar 10 at 15:31
  

  
  
  
  

add a comment | 

$begingroup$

Let $f: mathbb{R} to mathbb{R}$ be continuous and $g: mathbb{R} to mathbb{R}$ Lebesgue measurable. Is $f circ g$ Borel measurable?

I think that the answer to this question is no. I know that the hypothesis implies that $f circ g$ is Lebesgue measurable, but I am sure that there is a counter-example to show that $f circ g$ is not Borel measurable, since the Borel sigma algebra is a proper subset of the Lebesgue sigma algebra.

Note: In my textbook, a function is measurable with respect to a sigma algebra $mathcal{E}$ if and only if $[f > alpha] = {x : f(x) > alpha} in mathcal{E}$ for all $alpha in mathbb{R}$.

My current idea is to find a Lebesgue measurable set $D$ that is not Borel measurable. Then define $g(x) = chi_{_{D}}$, which is Lebesgue measurable, as $D$ is Lebesgue measurable. Then $f(x) = x$ is continuous and $f circ g = g$, but $g$ is not Borel measurable.

measure-theory

share|cite|improve this question

edited Mar 10 at 3:05

johnny133253

asked Mar 10 at 2:02

johnny133253johnny133253

463111

$endgroup$

share|cite|improve this question

edited Mar 10 at 3:05

johnny133253

asked Mar 10 at 2:02

johnny133253johnny133253

463111

share|cite|improve this question

edited Mar 10 at 3:05

johnny133253

asked Mar 10 at 2:02

johnny133253johnny133253

463111

asked Mar 10 at 2:02

johnny133253johnny133253

463111

asked Mar 10 at 2:02

johnny133253johnny133253

463111