Help in understanding the proof of a claim made in the proof of the Beppi Levi Theorem in Schilling.
$begingroup$
In the proof of the Beppo Levi theorem in Schilling the following claim is made:
$$fleq uRightarrow I_mu(f)leqsup_{jinmathbb{N}}int u_j dmu.$$
The proof of this claim is not hard but there is one point that I have a hard time seeing of why it is done, which is why is the parameter $alpha$ being introduced.
Let $(X,mathcal{A},mu)$ be a measure space,
$f=sumlimits_{k=0}^My_kmathbf{1}_{A_k}$ a standard representation of a non-negative measurable simple function, $u_i:Xrightarrow [0,infty]$ an increasing sequence of measurable functions and $u=suplimits_{iinmathbb{N}}u_i$. Define,
$I_mu(f):=sumlimits_{k=0}^My_kmu(A_k)$ and for any non-negative measurable function $w$,
$int wdmu:=sup{I_mu(g):gleq w$ & $ggeq 0$ measurable simple function$}.$
The proof of the claim from Schilling is as follows. For a simple non-negative $f$ such that $fleq u$, we can see that
$$(forallalphain(0,1))(forall xin X)(exists N(alpha,x))(forall jin mathbb{N})(jgeq N(alpha,x)Rightarrow alpha f(x)leq u_j(x))spacespace space -(dagger)$$
Thus the (measurable) sets $B_j:={alpha f leq u_j}uparrow X$ as $juparrowinfty$. Also we clearly have, $$alpha fmathbf{1}_{B_j}fleq mathbf{1}_{B_j}u_jleq u_j.$$
So then,
$$sum_{k=0}^Malpha y_kmu(A_kcap B_j)=I_mubigg(sum_{k=0}^Malpha y_kmathbf{1}_{A_k}mathbf{1}_{B_j}bigg)=I_mu(alphamathbf{1}_{B_j}f)leqint u_jdmuleqsup_{jinmathbb{N}}int u_jdmu.$$
Then by $sigma$-additivity of $mu$, $mu(A_kcap B_j)uparrow mu(A_k)$ as $B_juparrow X$, $juparrow infty$.
Thus $$alpha I_mu(f)=alphasumlimits_{k=0}^My_kmu(A_k)leqsup_{jinmathbb{N}}int u_jdmu$$ and letting $alpharightarrow 1$, the claim follows.
Now my problem is with $(dagger)$ and why the $alpha$ is needed? Now as $alphain(0,1)$ and $fgeq 0$, we obviously have $alpha fleq f$. So if $fleq u$ we then have $alpha fleq sup_{iinmathbb{N}}u_i$, and consequently for a $xin X$ there is a $N(alpha,x)inmathbb{N}$, such that $alpha f(x)leq u_{N(alpha,c)}(x)$ and as the $u_i$'s are increasing we have that $alpha f(x)leq u_j(x)$, for all $jgeq N(alpha,x)$.
But from this reasoning and the one above it will seem to work if the $alpha$ was never introduced, so I would appreciate any help to explain to me why I am wrong in thinking the $alpha$ is not needed.
Thanks in advance.
real-analysis measure-theory lebesgue-integral
asked Jan 16 at 21:42
user152874user152874
21719
$endgroup$
add a comment |
$begingroup$
In the proof of the Beppo Levi theorem in Schilling the following claim is made:
$$fleq uRightarrow I_mu(f)leqsup_{jinmathbb{N}}int u_j dmu.$$
The proof of this claim is not hard but there is one point that I have a hard time seeing of why it is done, which is why is the parameter $alpha$ being introduced.
Let $(X,mathcal{A},mu)$ be a measure space,
$f=sumlimits_{k=0}^My_kmathbf{1}_{A_k}$ a standard representation of a non-negative measurable simple function, $u_i:Xrightarrow [0,infty]$ an increasing sequence of measurable functions and $u=suplimits_{iinmathbb{N}}u_i$. Define,
$I_mu(f):=sumlimits_{k=0}^My_kmu(A_k)$ and for any non-negative measurable function $w$,
$int wdmu:=sup{I_mu(g):gleq w$ & $ggeq 0$ measurable simple function$}.$
The proof of the claim from Schilling is as follows. For a simple non-negative $f$ such that $fleq u$, we can see that
$$(forallalphain(0,1))(forall xin X)(exists N(alpha,x))(forall jin mathbb{N})(jgeq N(alpha,x)Rightarrow alpha f(x)leq u_j(x))spacespace space -(dagger)$$
Thus the (measurable) sets $B_j:={alpha f leq u_j}uparrow X$ as $juparrowinfty$. Also we clearly have, $$alpha fmathbf{1}_{B_j}fleq mathbf{1}_{B_j}u_jleq u_j.$$
So then,
$$sum_{k=0}^Malpha y_kmu(A_kcap B_j)=I_mubigg(sum_{k=0}^Malpha y_kmathbf{1}_{A_k}mathbf{1}_{B_j}bigg)=I_mu(alphamathbf{1}_{B_j}f)leqint u_jdmuleqsup_{jinmathbb{N}}int u_jdmu.$$
Then by $sigma$-additivity of $mu$, $mu(A_kcap B_j)uparrow mu(A_k)$ as $B_juparrow X$, $juparrow infty$.
Thus $$alpha I_mu(f)=alphasumlimits_{k=0}^My_kmu(A_k)leqsup_{jinmathbb{N}}int u_jdmu$$ and letting $alpharightarrow 1$, the claim follows.
Now my problem is with $(dagger)$ and why the $alpha$ is needed? Now as $alphain(0,1)$ and $fgeq 0$, we obviously have $alpha fleq f$. So if $fleq u$ we then have $alpha fleq sup_{iinmathbb{N}}u_i$, and consequently for a $xin X$ there is a $N(alpha,x)inmathbb{N}$, such that $alpha f(x)leq u_{N(alpha,c)}(x)$ and as the $u_i$'s are increasing we have that $alpha f(x)leq u_j(x)$, for all $jgeq N(alpha,x)$.
But from this reasoning and the one above it will seem to work if the $alpha$ was never introduced, so I would appreciate any help to explain to me why I am wrong in thinking the $alpha$ is not needed.
Thanks in advance.
real-analysis measure-theory lebesgue-integral
asked Jan 16 at 21:42
user152874user152874
21719
$endgroup$
3
$begingroup$
$f leq sup_i u_i$ does not imply $f leq u_{N(alpha,x)}(x)$ ... just consider e.g. $f=1$ and $u_i := 1-1/i$. That's why $alpha$ is needed, we cannot simply put $alpha=1$.
$endgroup$
– saz
Jan 16 at 21:55
$begingroup$
Thanks, I completely missed the case that $f=sup_iu_i$
$endgroup$
– user152874
Jan 17 at 7:10
add a comment |
$begingroup$
In the proof of the Beppo Levi theorem in Schilling the following claim is made:
$$fleq uRightarrow I_mu(f)leqsup_{jinmathbb{N}}int u_j dmu.$$
The proof of this claim is not hard but there is one point that I have a hard time seeing of why it is done, which is why is the parameter $alpha$ being introduced.
Let $(X,mathcal{A},mu)$ be a measure space,
$f=sumlimits_{k=0}^My_kmathbf{1}_{A_k}$ a standard representation of a non-negative measurable simple function, $u_i:Xrightarrow [0,infty]$ an increasing sequence of measurable functions and $u=suplimits_{iinmathbb{N}}u_i$. Define,
$I_mu(f):=sumlimits_{k=0}^My_kmu(A_k)$ and for any non-negative measurable function $w$,
$int wdmu:=sup{I_mu(g):gleq w$ & $ggeq 0$ measurable simple function$}.$
The proof of the claim from Schilling is as follows. For a simple non-negative $f$ such that $fleq u$, we can see that
$$(forallalphain(0,1))(forall xin X)(exists N(alpha,x))(forall jin mathbb{N})(jgeq N(alpha,x)Rightarrow alpha f(x)leq u_j(x))spacespace space -(dagger)$$
Thus the (measurable) sets $B_j:={alpha f leq u_j}uparrow X$ as $juparrowinfty$. Also we clearly have, $$alpha fmathbf{1}_{B_j}fleq mathbf{1}_{B_j}u_jleq u_j.$$
So then,
$$sum_{k=0}^Malpha y_kmu(A_kcap B_j)=I_mubigg(sum_{k=0}^Malpha y_kmathbf{1}_{A_k}mathbf{1}_{B_j}bigg)=I_mu(alphamathbf{1}_{B_j}f)leqint u_jdmuleqsup_{jinmathbb{N}}int u_jdmu.$$
Then by $sigma$-additivity of $mu$, $mu(A_kcap B_j)uparrow mu(A_k)$ as $B_juparrow X$, $juparrow infty$.
Thus $$alpha I_mu(f)=alphasumlimits_{k=0}^My_kmu(A_k)leqsup_{jinmathbb{N}}int u_jdmu$$ and letting $alpharightarrow 1$, the claim follows.
Now my problem is with $(dagger)$ and why the $alpha$ is needed? Now as $alphain(0,1)$ and $fgeq 0$, we obviously have $alpha fleq f$. So if $fleq u$ we then have $alpha fleq sup_{iinmathbb{N}}u_i$, and consequently for a $xin X$ there is a $N(alpha,x)inmathbb{N}$, such that $alpha f(x)leq u_{N(alpha,c)}(x)$ and as the $u_i$'s are increasing we have that $alpha f(x)leq u_j(x)$, for all $jgeq N(alpha,x)$.
But from this reasoning and the one above it will seem to work if the $alpha$ was never introduced, so I would appreciate any help to explain to me why I am wrong in thinking the $alpha$ is not needed.
Thanks in advance.
real-analysis measure-theory lebesgue-integral
asked Jan 16 at 21:42
user152874user152874
21719
$endgroup$
In the proof of the Beppo Levi theorem in Schilling the following claim is made:
$$fleq uRightarrow I_mu(f)leqsup_{jinmathbb{N}}int u_j dmu.$$
The proof of this claim is not hard but there is one point that I have a hard time seeing of why it is done, which is why is the parameter $alpha$ being introduced.
Let $(X,mathcal{A},mu)$ be a measure space,
$f=sumlimits_{k=0}^My_kmathbf{1}_{A_k}$ a standard representation of a non-negative measurable simple function, $u_i:Xrightarrow [0,infty]$ an increasing sequence of measurable functions and $u=suplimits_{iinmathbb{N}}u_i$. Define,
$I_mu(f):=sumlimits_{k=0}^My_kmu(A_k)$ and for any non-negative measurable function $w$,
$int wdmu:=sup{I_mu(g):gleq w$ & $ggeq 0$ measurable simple function$}.$
The proof of the claim from Schilling is as follows. For a simple non-negative $f$ such that $fleq u$, we can see that
$$(forallalphain(0,1))(forall xin X)(exists N(alpha,x))(forall jin mathbb{N})(jgeq N(alpha,x)Rightarrow alpha f(x)leq u_j(x))spacespace space -(dagger)$$
Thus the (measurable) sets $B_j:={alpha f leq u_j}uparrow X$ as $juparrowinfty$. Also we clearly have, $$alpha fmathbf{1}_{B_j}fleq mathbf{1}_{B_j}u_jleq u_j.$$
So then,
$$sum_{k=0}^Malpha y_kmu(A_kcap B_j)=I_mubigg(sum_{k=0}^Malpha y_kmathbf{1}_{A_k}mathbf{1}_{B_j}bigg)=I_mu(alphamathbf{1}_{B_j}f)leqint u_jdmuleqsup_{jinmathbb{N}}int u_jdmu.$$
Then by $sigma$-additivity of $mu$, $mu(A_kcap B_j)uparrow mu(A_k)$ as $B_juparrow X$, $juparrow infty$.
Thus $$alpha I_mu(f)=alphasumlimits_{k=0}^My_kmu(A_k)leqsup_{jinmathbb{N}}int u_jdmu$$ and letting $alpharightarrow 1$, the claim follows.
Now my problem is with $(dagger)$ and why the $alpha$ is needed? Now as $alphain(0,1)$ and $fgeq 0$, we obviously have $alpha fleq f$. So if $fleq u$ we then have $alpha fleq sup_{iinmathbb{N}}u_i$, and consequently for a $xin X$ there is a $N(alpha,x)inmathbb{N}$, such that $alpha f(x)leq u_{N(alpha,c)}(x)$ and as the $u_i$'s are increasing we have that $alpha f(x)leq u_j(x)$, for all $jgeq N(alpha,x)$.
But from this reasoning and the one above it will seem to work if the $alpha$ was never introduced, so I would appreciate any help to explain to me why I am wrong in thinking the $alpha$ is not needed.
Thanks in advance.
real-analysis measure-theory lebesgue-integral
real-analysis measure-theory lebesgue-integral
asked Jan 16 at 21:42
user152874user152874
21719
asked Jan 16 at 21:42
user152874user152874
21719
asked Jan 16 at 21:42
user152874user152874
21719
asked Jan 16 at 21:42
user152874user152874
21719
asked Jan 16 at 21:42
user152874user152874
21719
21719
3
$begingroup$
$f leq sup_i u_i$ does not imply $f leq u_{N(alpha,x)}(x)$ ... just consider e.g. $f=1$ and $u_i := 1-1/i$. That's why $alpha$ is needed, we cannot simply put $alpha=1$.
$endgroup$
– saz
Jan 16 at 21:55
$begingroup$
Thanks, I completely missed the case that $f=sup_iu_i$
$endgroup$
– user152874
Jan 17 at 7:10
add a comment |
3
$begingroup$
$f leq sup_i u_i$ does not imply $f leq u_{N(alpha,x)}(x)$ ... just consider e.g. $f=1$ and $u_i := 1-1/i$. That's why $alpha$ is needed, we cannot simply put $alpha=1$.
$endgroup$
– saz
Jan 16 at 21:55
$begingroup$
Thanks, I completely missed the case that $f=sup_iu_i$
$endgroup$
– user152874
Jan 17 at 7:10
3
3
$begingroup$
$f leq sup_i u_i$ does not imply $f leq u_{N(alpha,x)}(x)$ ... just consider e.g. $f=1$ and $u_i := 1-1/i$. That's why $alpha$ is needed, we cannot simply put $alpha=1$.
$endgroup$
– saz
Jan 16 at 21:55
$begingroup$
$f leq sup_i u_i$ does not imply $f leq u_{N(alpha,x)}(x)$ ... just consider e.g. $f=1$ and $u_i := 1-1/i$. That's why $alpha$ is needed, we cannot simply put $alpha=1$.
$endgroup$
– saz
Jan 16 at 21:55
$begingroup$
Thanks, I completely missed the case that $f=sup_iu_i$
$endgroup$
– user152874
Jan 17 at 7:10
$begingroup$
Thanks, I completely missed the case that $f=sup_iu_i$
$endgroup$
– user152874
Jan 17 at 7:10
add a comment |
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3
$begingroup$
$f leq sup_i u_i$ does not imply $f leq u_{N(alpha,x)}(x)$ ... just consider e.g. $f=1$ and $u_i := 1-1/i$. That's why $alpha$ is needed, we cannot simply put $alpha=1$.
$endgroup$
– saz
Jan 16 at 21:55
$begingroup$
Thanks, I completely missed the case that $f=sup_iu_i$
$endgroup$
– user152874
Jan 17 at 7:10