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Integration w.r.t. pushforward measure

Radon–Nikodym derivative and “normal” derivativeMeasurable with respect to the product measureDoes the pushforward operator (on measures) preserve surjectiveness?$mu$ is a finite Borel measure on $Bbb R$, absolutely continuous w.r.t. to the Lebesgue measure $m$. Prove that $x mapsto mu(A+x)$ is continuous.Show that the measure is absolutely continuous w.r.t Lebesgue measureAre self-similar measures a.c. with respect to Lebesgue measure?The Radon Nikodym derivative is non-zero almost everywhereObtaining any measure as the pushforward of the uniform under a measurable map?Does Radon-Nikodym imply Riesz Representation Theorem?Construction of a “density” or a Radon-Nikodym for a Semicontinuous Distribution

0

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Problem: Let $phi colon [0,1] to [0,1]$ be a continuous function and let $mu$ be a Borel probability measure on $[0,1]$. Suppose $mu(phi^{-1}(E)) = 0$ for every Borel set $E subseteq [0,1]$ with $mu(E) = 0$. Show that there is a Borel measurable function $g colon [0,1] to [0,infty)$ such that
$$int_0^1 f(phi(x)) hspace{1mm} dmu(x) = int_0^1 f(x)g(x) hspace{1mm} dmu(x)$$
for all continuous functions $f colon [0,1] to mathbb{R}$.

I am a bit confused by the hypotheses in the problem statement. It seems to be encouraging you to use the Riesz-representation theorem, but it seems unnecessary to assume that $phi$ is continuous as opposed to merely measurable, and the desired equality should be true for all measurable $f colon [0,1] to mathbb{R}$. The left-hand integral is just the integral of $f$ with respect to the pushforward measure $nu(E) := mu(phi^{-1}(E))$ and the hypothesis that $mu(phi^{-1}(E)) = 0$ if $mu(E) = 0$ is precisely to say that $nu$ is absolutely continuous with respect to $mu$. Therefore we have
$$int_0^1 f hspace{1mm} dnu(x) = int_0^1 f(x)h(x) hspace{1mm} dmu(x),$$
where $h$ is the Radon-Nikodym derivative of $nu$ with respect to $mu$. Since $mu$ is a nonnegative measure, so is $nu$, thus $h geq 0$.

real-analysis measure-theory

share|cite|improve this question

asked Mar 19 at 5:52

Ethan AlwaiseEthan Alwaise

6,471717

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You are absolutely right. Continuity is too strong an assumption in this measure theoretic context. However note that the integrals may not exist for all measurable functions, so you need bounded measurable functions.
  $endgroup$
  – Kavi Rama Murthy
  Mar 19 at 6:16
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Problem: Let $phi colon [0,1] to [0,1]$ be a continuous function and let $mu$ be a Borel probability measure on $[0,1]$. Suppose $mu(phi^{-1}(E)) = 0$ for every Borel set $E subseteq [0,1]$ with $mu(E) = 0$. Show that there is a Borel measurable function $g colon [0,1] to [0,infty)$ such that
$$int_0^1 f(phi(x)) hspace{1mm} dmu(x) = int_0^1 f(x)g(x) hspace{1mm} dmu(x)$$
for all continuous functions $f colon [0,1] to mathbb{R}$.

I am a bit confused by the hypotheses in the problem statement. It seems to be encouraging you to use the Riesz-representation theorem, but it seems unnecessary to assume that $phi$ is continuous as opposed to merely measurable, and the desired equality should be true for all measurable $f colon [0,1] to mathbb{R}$. The left-hand integral is just the integral of $f$ with respect to the pushforward measure $nu(E) := mu(phi^{-1}(E))$ and the hypothesis that $mu(phi^{-1}(E)) = 0$ if $mu(E) = 0$ is precisely to say that $nu$ is absolutely continuous with respect to $mu$. Therefore we have
$$int_0^1 f hspace{1mm} dnu(x) = int_0^1 f(x)h(x) hspace{1mm} dmu(x),$$
where $h$ is the Radon-Nikodym derivative of $nu$ with respect to $mu$. Since $mu$ is a nonnegative measure, so is $nu$, thus $h geq 0$.

real-analysis measure-theory

share|cite|improve this question

asked Mar 19 at 5:52

Ethan AlwaiseEthan Alwaise

6,471717

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You are absolutely right. Continuity is too strong an assumption in this measure theoretic context. However note that the integrals may not exist for all measurable functions, so you need bounded measurable functions.
  $endgroup$
  – Kavi Rama Murthy
  Mar 19 at 6:16
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Problem: Let $phi colon [0,1] to [0,1]$ be a continuous function and let $mu$ be a Borel probability measure on $[0,1]$. Suppose $mu(phi^{-1}(E)) = 0$ for every Borel set $E subseteq [0,1]$ with $mu(E) = 0$. Show that there is a Borel measurable function $g colon [0,1] to [0,infty)$ such that
$$int_0^1 f(phi(x)) hspace{1mm} dmu(x) = int_0^1 f(x)g(x) hspace{1mm} dmu(x)$$
for all continuous functions $f colon [0,1] to mathbb{R}$.

I am a bit confused by the hypotheses in the problem statement. It seems to be encouraging you to use the Riesz-representation theorem, but it seems unnecessary to assume that $phi$ is continuous as opposed to merely measurable, and the desired equality should be true for all measurable $f colon [0,1] to mathbb{R}$. The left-hand integral is just the integral of $f$ with respect to the pushforward measure $nu(E) := mu(phi^{-1}(E))$ and the hypothesis that $mu(phi^{-1}(E)) = 0$ if $mu(E) = 0$ is precisely to say that $nu$ is absolutely continuous with respect to $mu$. Therefore we have
$$int_0^1 f hspace{1mm} dnu(x) = int_0^1 f(x)h(x) hspace{1mm} dmu(x),$$
where $h$ is the Radon-Nikodym derivative of $nu$ with respect to $mu$. Since $mu$ is a nonnegative measure, so is $nu$, thus $h geq 0$.

real-analysis measure-theory

share|cite|improve this question

asked Mar 19 at 5:52

Ethan AlwaiseEthan Alwaise

6,471717

$endgroup$

share|cite|improve this question

asked Mar 19 at 5:52

Ethan AlwaiseEthan Alwaise

6,471717

share|cite|improve this question

asked Mar 19 at 5:52

Ethan AlwaiseEthan Alwaise

6,471717

asked Mar 19 at 5:52

Ethan AlwaiseEthan Alwaise

6,471717

asked Mar 19 at 5:52

Ethan AlwaiseEthan Alwaise

6,471717