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Proof about Borel measurable function

Cartesian Product of Borel Sets is Borel AgainShow that continuous functions on $mathbb R$ are Borel-measurableLebesgue and Borel MeasurableBorel-set, open, measurable function.Measurability of product of Borel measurable functions with different domains?Deriving a mappingWhy do we use the Borel sigma algebra for the codomain of a measurable function?Rudin's definition on measurable functionComposition of Borel-measurable and continuous functionsComposition Borel Measurable Function with Measurable FunctionBorel measurable function $f : X rightarrow mathbbC$.Projection of a measurable cylinder of the product $sigma$-algebra

0

$begingroup$

Let assume that the $X = [x_1,1,x_1,2]times[x_2,1,x_3,2]times[x_3,1,x_3,2]times[x_4,1,x_4,2]$, $Y=[y_1,1,y_1,2]$ where $x_n,m $ and $ y_n,m$ are real numbers for all $n,m$.

Show that $f: X to Y $ is Borel measurable function if $f$ is continuous.

I think the domain of the function $f$ is a Borel set because the $X$ is a cartesian product of Borel sets, Cartesian Product of Borel Sets is Borel Again.

And I know that continuous functions are measurable w.r.t. Borel sigma-algebra. But I want to clear proof for the problem.

real-analysis borel-measures

share|cite|improve this question

edited Mar 21 at 2:45

Andrews

1,2812422

asked Mar 20 at 20:10

Heasung KimHeasung Kim

574

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you check this question?
  $endgroup$
  – dcolazin
  Mar 20 at 20:54
  

  
  
  
  

add a comment | 

0

$begingroup$

Let assume that the $X = [x_1,1,x_1,2]times[x_2,1,x_3,2]times[x_3,1,x_3,2]times[x_4,1,x_4,2]$, $Y=[y_1,1,y_1,2]$ where $x_n,m $ and $ y_n,m$ are real numbers for all $n,m$.

Show that $f: X to Y $ is Borel measurable function if $f$ is continuous.

I think the domain of the function $f$ is a Borel set because the $X$ is a cartesian product of Borel sets, Cartesian Product of Borel Sets is Borel Again.

And I know that continuous functions are measurable w.r.t. Borel sigma-algebra. But I want to clear proof for the problem.

real-analysis borel-measures

share|cite|improve this question

edited Mar 21 at 2:45

Andrews

1,2812422

asked Mar 20 at 20:10

Heasung KimHeasung Kim

574

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you check this question?
  $endgroup$
  – dcolazin
  Mar 20 at 20:54
  

  
  
  
  

add a comment | 

$begingroup$

Let assume that the $X = [x_1,1,x_1,2]times[x_2,1,x_3,2]times[x_3,1,x_3,2]times[x_4,1,x_4,2]$, $Y=[y_1,1,y_1,2]$ where $x_n,m $ and $ y_n,m$ are real numbers for all $n,m$.

Show that $f: X to Y $ is Borel measurable function if $f$ is continuous.

I think the domain of the function $f$ is a Borel set because the $X$ is a cartesian product of Borel sets, Cartesian Product of Borel Sets is Borel Again.

And I know that continuous functions are measurable w.r.t. Borel sigma-algebra. But I want to clear proof for the problem.

real-analysis borel-measures

share|cite|improve this question

edited Mar 21 at 2:45

Andrews

1,2812422

asked Mar 20 at 20:10

Heasung KimHeasung Kim

574

$endgroup$

share|cite|improve this question

edited Mar 21 at 2:45

Andrews

1,2812422

asked Mar 20 at 20:10

Heasung KimHeasung Kim

574

share|cite|improve this question

edited Mar 21 at 2:45

Andrews

1,2812422

asked Mar 20 at 20:10

Heasung KimHeasung Kim

574

edited Mar 21 at 2:45

Andrews

1,2812422

edited Mar 21 at 2:45

Andrews

1,2812422

asked Mar 20 at 20:10

Heasung KimHeasung Kim

574

asked Mar 20 at 20:10

Heasung KimHeasung Kim

574