Fdtxjr

Can somebody explain Brexit in a few child-proof sentences?

In Star Trek IV, why did the Bounty go back to a time when whales were already rare?

What should I use for Mishna study?

Is there any significance to the Valyrian Stone vault door of Qarth?

Books on the History of math research at European universities

Resetting two CD4017 counters simultaneously, only one resets

How to prevent YouTube from showing already watched videos?

Organic chemistry Iodoform Reaction

How can I raise concerns with a new DM about XP splitting?

Why are on-board computers allowed to change controls without notifying the pilots?

Is there enough fresh water in the world to eradicate the drinking water crisis?

Could solar power be utilized and substitute coal in the 19th century?

Identify a stage play about a VR experience in which participants are encouraged to simulate performing horrific activities

Bob has never been a M before

Who must act to prevent Brexit on March 29th?

Invariance of results when scaling explanatory variables in logistic regression, is there a proof?

Giant Toughroad SLR 2 for 200 miles in two days, will it make it?

How will losing mobility of one hand affect my career as a programmer?

Lifted its hind leg on or lifted its hind leg towards?

Is the next prime number always the next number divisible by the current prime number, except for any numbers previously divisible by primes?

What (else) happened July 1st 1858 in London?

Can I create an upright 7-foot × 5-foot wall with the Minor Illusion spell?

Hostile work environment after whistle-blowing on coworker and our boss. What do I do?

Blender - show edges angles “direction”

How to prove Dilation property of Lebesgue integral

If $f$ is Lebesgue measurable, prove that there is a Borel measurable function $g$ such that $f=g$ except, possibly, on a Borel set of measure zero.Help with a Lebesgue integration problem.How to show $int_[0, +infty) frac21+x^2 dx$ Lebesgue integrable?How to prove an inequality of Lebesgue integral?Questions of an exercise in Lebesgue integralDilation of Real Valued Lebesgue IntegralProve that lebesgue integrable equal lebesgue measureIf $f in L^+$ and $int f<infty$ then there exists a null setUsing the Lebesgue dominated convergence theoremLebesgue-integrability of the Dirac delta function?

0

$begingroup$

Let $fin L^1(mathbbR^d), a_1,dots,a_d>0$, and $a=(a_1,dots,a_d)$. Define
$$g(x)=f(a_1^-1x_1,dots,a_d^-1x_d).$$
Show that $din L^1(mathbb R^d)$ and that $$int g=left(prod^d_j=1a_jright)int f.$$

$textbfMy Attempt:$ Since $f$ is integrable, it is also measurable, and hence there is an increasing sequence of simple functions $(varphi_n)_n$, such that $varphi_nto f$ a.e. This implies that $varphi_n(a^-1x)to f(a^-1x)$ a.e, where $a^-1xequiv (a_1^-1x_1,dots,a_d^-1x_d).$ This implies that $g$ is measurable since it is a limit of measurable functions.

Let $varphi(x)=sum_j=1^Nc_jcdot1_E_j(x)$ be a simple function, where the $E_j$ are measurable sets. Then we have by dilation invariance of the Lebesgue measure and linearity of the Lebesgue integral we get for $psi(x)=varphi(a^-1x)$
$$largeintpsi=sum^N_j=1c_km(a^-1E_j)=prod^d_j=1a_jsum^N_j=1c_k m(E_j).$$
For a non-negative integrable function, the monotone convergence theorem comes to the rescue, since we can approximate our function by an increasing sequence of simple functions, and then the result follows from the above.

Here is the work. Let $g$ be as at the start, but non-negative for simplicity. Then we have by the remark, using the Monotone convergence theorem that
$$largeint g=limlimits_ntoinftyintvarphi_n(a^-1x)=prod^d_j=1a_jlimlimits_ntoinftyintvarphi_n=prod^d_j=1a_jint f.$$

---

Is my work above correct? Any feedback is much welcomed.

Thank you for your time.

real-analysis proof-verification lebesgue-integral lebesgue-measure

share|cite|improve this question

edited Mar 20 at 9:36

Gaby Alfonso

asked Mar 16 at 20:31

Gaby AlfonsoGaby Alfonso

1,1811318

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $m([0,2]) = m(2[0, 1]) = 2m([0, 1]) = 2$, right? So something is shady in here. $int varphi_n(a^-1x)dx = sum c_jm(aE_j)$, $1_E(a^-1x) = 1_aE(x)$
  $endgroup$
  – Jakobian
  Mar 16 at 20:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jakobian My bad, I worked out that $1_E(ax)=1_a^-1E(x)$ for $mathbbR$ and then we extend this to $mathbbR^d$ by using the product measure, since if $E=E_1timescdotstimes E_d$, then $m(a^-1E)=prod^d_j=1m(a_j^-1E_j)=prod^d_j=1a_j^-1m(E)$. Is that right?
  $endgroup$
  – Gaby Alfonso
  Mar 16 at 21:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that's right
  $endgroup$
  – Jakobian
  Mar 16 at 22:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not dilation invariance. It's the dilation property of the Lebesgue integral in this setting.
  $endgroup$
  – zhw.
  Mar 17 at 15:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @zhw. My mistake, sorry. I corrected the error.
  $endgroup$
  – Gaby Alfonso
  Mar 20 at 9:38
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $fin L^1(mathbbR^d), a_1,dots,a_d>0$, and $a=(a_1,dots,a_d)$. Define
$$g(x)=f(a_1^-1x_1,dots,a_d^-1x_d).$$
Show that $din L^1(mathbb R^d)$ and that $$int g=left(prod^d_j=1a_jright)int f.$$

$textbfMy Attempt:$ Since $f$ is integrable, it is also measurable, and hence there is an increasing sequence of simple functions $(varphi_n)_n$, such that $varphi_nto f$ a.e. This implies that $varphi_n(a^-1x)to f(a^-1x)$ a.e, where $a^-1xequiv (a_1^-1x_1,dots,a_d^-1x_d).$ This implies that $g$ is measurable since it is a limit of measurable functions.

Let $varphi(x)=sum_j=1^Nc_jcdot1_E_j(x)$ be a simple function, where the $E_j$ are measurable sets. Then we have by dilation invariance of the Lebesgue measure and linearity of the Lebesgue integral we get for $psi(x)=varphi(a^-1x)$
$$largeintpsi=sum^N_j=1c_km(a^-1E_j)=prod^d_j=1a_jsum^N_j=1c_k m(E_j).$$
For a non-negative integrable function, the monotone convergence theorem comes to the rescue, since we can approximate our function by an increasing sequence of simple functions, and then the result follows from the above.

Here is the work. Let $g$ be as at the start, but non-negative for simplicity. Then we have by the remark, using the Monotone convergence theorem that
$$largeint g=limlimits_ntoinftyintvarphi_n(a^-1x)=prod^d_j=1a_jlimlimits_ntoinftyintvarphi_n=prod^d_j=1a_jint f.$$

---

Is my work above correct? Any feedback is much welcomed.

Thank you for your time.

real-analysis proof-verification lebesgue-integral lebesgue-measure

share|cite|improve this question

edited Mar 20 at 9:36

Gaby Alfonso

asked Mar 16 at 20:31

Gaby AlfonsoGaby Alfonso

1,1811318

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $m([0,2]) = m(2[0, 1]) = 2m([0, 1]) = 2$, right? So something is shady in here. $int varphi_n(a^-1x)dx = sum c_jm(aE_j)$, $1_E(a^-1x) = 1_aE(x)$
  $endgroup$
  – Jakobian
  Mar 16 at 20:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jakobian My bad, I worked out that $1_E(ax)=1_a^-1E(x)$ for $mathbbR$ and then we extend this to $mathbbR^d$ by using the product measure, since if $E=E_1timescdotstimes E_d$, then $m(a^-1E)=prod^d_j=1m(a_j^-1E_j)=prod^d_j=1a_j^-1m(E)$. Is that right?
  $endgroup$
  – Gaby Alfonso
  Mar 16 at 21:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that's right
  $endgroup$
  – Jakobian
  Mar 16 at 22:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not dilation invariance. It's the dilation property of the Lebesgue integral in this setting.
  $endgroup$
  – zhw.
  Mar 17 at 15:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @zhw. My mistake, sorry. I corrected the error.
  $endgroup$
  – Gaby Alfonso
  Mar 20 at 9:38
  

  
  
  
  

add a comment | 

$begingroup$

Let $fin L^1(mathbbR^d), a_1,dots,a_d>0$, and $a=(a_1,dots,a_d)$. Define
$$g(x)=f(a_1^-1x_1,dots,a_d^-1x_d).$$
Show that $din L^1(mathbb R^d)$ and that $$int g=left(prod^d_j=1a_jright)int f.$$

$textbfMy Attempt:$ Since $f$ is integrable, it is also measurable, and hence there is an increasing sequence of simple functions $(varphi_n)_n$, such that $varphi_nto f$ a.e. This implies that $varphi_n(a^-1x)to f(a^-1x)$ a.e, where $a^-1xequiv (a_1^-1x_1,dots,a_d^-1x_d).$ This implies that $g$ is measurable since it is a limit of measurable functions.

Let $varphi(x)=sum_j=1^Nc_jcdot1_E_j(x)$ be a simple function, where the $E_j$ are measurable sets. Then we have by dilation invariance of the Lebesgue measure and linearity of the Lebesgue integral we get for $psi(x)=varphi(a^-1x)$
$$largeintpsi=sum^N_j=1c_km(a^-1E_j)=prod^d_j=1a_jsum^N_j=1c_k m(E_j).$$
For a non-negative integrable function, the monotone convergence theorem comes to the rescue, since we can approximate our function by an increasing sequence of simple functions, and then the result follows from the above.

Here is the work. Let $g$ be as at the start, but non-negative for simplicity. Then we have by the remark, using the Monotone convergence theorem that
$$largeint g=limlimits_ntoinftyintvarphi_n(a^-1x)=prod^d_j=1a_jlimlimits_ntoinftyintvarphi_n=prod^d_j=1a_jint f.$$

---

Is my work above correct? Any feedback is much welcomed.

Thank you for your time.

real-analysis proof-verification lebesgue-integral lebesgue-measure

share|cite|improve this question

edited Mar 20 at 9:36

Gaby Alfonso

asked Mar 16 at 20:31

Gaby AlfonsoGaby Alfonso

1,1811318

$endgroup$

share|cite|improve this question

edited Mar 20 at 9:36

Gaby Alfonso

asked Mar 16 at 20:31

Gaby AlfonsoGaby Alfonso

1,1811318

share|cite|improve this question

edited Mar 20 at 9:36

Gaby Alfonso

asked Mar 16 at 20:31

Gaby AlfonsoGaby Alfonso

1,1811318

asked Mar 16 at 20:31

Gaby AlfonsoGaby Alfonso

1,1811318

asked Mar 16 at 20:31

Gaby AlfonsoGaby Alfonso

1,1811318