Mixing
Varying definition of monotone convergence
$begingroup$
My current lecture states the theorem of monotone convergence:
Let $(X, mathcal{E}, mu)$ a measure space and $varphi_n: X
rightarrow mathbb{R}$ an increasing sequence of $mu$-integrable
functions with:
$$exists M in mathbb{R}: forall n in mathbb{N}:
int_Xvarphi_n,dmu leq M$$
Then $varphi := lim varphi_n: X rightarrow overline{mathbb{R}}$
is $mu$-integrable with:
$$int_X varphi,dmu = lim_{n rightarrow infty} varphi_n,dmu$$
Note: A function $f: X rightarrow mathbb{R}$ is called $mu$-integrable iff $int_X|f|,dmu < infty$.
Wikipedia states:
Let $(X, mathcal{E}, mu)$ be a measure space. For a pointwise
non-decreasing sequence of $mathcal{E}$-measurable non-negative
functions $f_k: X rightarrow [0, infty]$ consider the pointwise
limit $$f := lim_{krightarrowinfty} f_k$$ Then $f$ is
$mathcal{E}$-measurable with:
$$ lim_{krightarrowinfty} int_X f_k, dmu = int_X f,dmu$$
Now I'm a little bit confused my lecture doesn't require the functions of the sequence to be non-negative but introduces the extra upper bound. What is the difference here?
The Wikipedia definition allows it to set the integral of the limit equal to the limit of the integrals. The lecture definition permits this only under the condition, that the sequence is bounded. Is this really required?
integration measure-theory lebesgue-integral lebesgue-measure
asked Jan 14 at 14:20
user7802048user7802048
382211
$endgroup$
add a comment |
$begingroup$
My current lecture states the theorem of monotone convergence:
Let $(X, mathcal{E}, mu)$ a measure space and $varphi_n: X
rightarrow mathbb{R}$ an increasing sequence of $mu$-integrable
functions with:
$$exists M in mathbb{R}: forall n in mathbb{N}:
int_Xvarphi_n,dmu leq M$$
Then $varphi := lim varphi_n: X rightarrow overline{mathbb{R}}$
is $mu$-integrable with:
$$int_X varphi,dmu = lim_{n rightarrow infty} varphi_n,dmu$$
Note: A function $f: X rightarrow mathbb{R}$ is called $mu$-integrable iff $int_X|f|,dmu < infty$.
Wikipedia states:
Let $(X, mathcal{E}, mu)$ be a measure space. For a pointwise
non-decreasing sequence of $mathcal{E}$-measurable non-negative
functions $f_k: X rightarrow [0, infty]$ consider the pointwise
limit $$f := lim_{krightarrowinfty} f_k$$ Then $f$ is
$mathcal{E}$-measurable with:
$$ lim_{krightarrowinfty} int_X f_k, dmu = int_X f,dmu$$
Now I'm a little bit confused my lecture doesn't require the functions of the sequence to be non-negative but introduces the extra upper bound. What is the difference here?
The Wikipedia definition allows it to set the integral of the limit equal to the limit of the integrals. The lecture definition permits this only under the condition, that the sequence is bounded. Is this really required?
integration measure-theory lebesgue-integral lebesgue-measure
asked Jan 14 at 14:20
user7802048user7802048
382211
$endgroup$
$begingroup$
Can $int_X varphi_n dmu = -infty$?
$endgroup$
– BigbearZzz
Jan 14 at 14:38
$begingroup$
@BigbearZzz No, since it's required that the $varphi_n$ are $mu$-integrable (the integral of the absolute value is not $infty$)
$endgroup$
– user7802048
Jan 14 at 15:07
$begingroup$
So you're using the word $mu$-integrable in the sense that $int_X |varphi| dmu < infty$? I just want to make sure because I've seen people use it differently.
$endgroup$
– BigbearZzz
Jan 14 at 15:13
$begingroup$
@BigbearZzz Yes exactly. I'll edit my question to include this information.
$endgroup$
– user7802048
Jan 14 at 15:15
add a comment |
$begingroup$
My current lecture states the theorem of monotone convergence:
Let $(X, mathcal{E}, mu)$ a measure space and $varphi_n: X
rightarrow mathbb{R}$ an increasing sequence of $mu$-integrable
functions with:
$$exists M in mathbb{R}: forall n in mathbb{N}:
int_Xvarphi_n,dmu leq M$$
Then $varphi := lim varphi_n: X rightarrow overline{mathbb{R}}$
is $mu$-integrable with:
$$int_X varphi,dmu = lim_{n rightarrow infty} varphi_n,dmu$$
Note: A function $f: X rightarrow mathbb{R}$ is called $mu$-integrable iff $int_X|f|,dmu < infty$.
Wikipedia states:
Let $(X, mathcal{E}, mu)$ be a measure space. For a pointwise
non-decreasing sequence of $mathcal{E}$-measurable non-negative
functions $f_k: X rightarrow [0, infty]$ consider the pointwise
limit $$f := lim_{krightarrowinfty} f_k$$ Then $f$ is
$mathcal{E}$-measurable with:
$$ lim_{krightarrowinfty} int_X f_k, dmu = int_X f,dmu$$
Now I'm a little bit confused my lecture doesn't require the functions of the sequence to be non-negative but introduces the extra upper bound. What is the difference here?
The Wikipedia definition allows it to set the integral of the limit equal to the limit of the integrals. The lecture definition permits this only under the condition, that the sequence is bounded. Is this really required?
integration measure-theory lebesgue-integral lebesgue-measure
asked Jan 14 at 14:20
user7802048user7802048
382211
$endgroup$
My current lecture states the theorem of monotone convergence:
Let $(X, mathcal{E}, mu)$ a measure space and $varphi_n: X
rightarrow mathbb{R}$ an increasing sequence of $mu$-integrable
functions with:
$$exists M in mathbb{R}: forall n in mathbb{N}:
int_Xvarphi_n,dmu leq M$$
Then $varphi := lim varphi_n: X rightarrow overline{mathbb{R}}$
is $mu$-integrable with:
$$int_X varphi,dmu = lim_{n rightarrow infty} varphi_n,dmu$$
Note: A function $f: X rightarrow mathbb{R}$ is called $mu$-integrable iff $int_X|f|,dmu < infty$.
Wikipedia states:
Let $(X, mathcal{E}, mu)$ be a measure space. For a pointwise
non-decreasing sequence of $mathcal{E}$-measurable non-negative
functions $f_k: X rightarrow [0, infty]$ consider the pointwise
limit $$f := lim_{krightarrowinfty} f_k$$ Then $f$ is
$mathcal{E}$-measurable with:
$$ lim_{krightarrowinfty} int_X f_k, dmu = int_X f,dmu$$
Now I'm a little bit confused my lecture doesn't require the functions of the sequence to be non-negative but introduces the extra upper bound. What is the difference here?
The Wikipedia definition allows it to set the integral of the limit equal to the limit of the integrals. The lecture definition permits this only under the condition, that the sequence is bounded. Is this really required?
integration measure-theory lebesgue-integral lebesgue-measure
integration measure-theory lebesgue-integral lebesgue-measure
asked Jan 14 at 14:20
user7802048user7802048
382211
asked Jan 14 at 14:20
user7802048user7802048
382211
edited Jan 14 at 15:17
user7802048
asked Jan 14 at 14:20
user7802048user7802048
382211
asked Jan 14 at 14:20
user7802048user7802048
382211
asked Jan 14 at 14:20
user7802048user7802048
382211
382211
$begingroup$
Can $int_X varphi_n dmu = -infty$?
$endgroup$
– BigbearZzz
Jan 14 at 14:38
$begingroup$
@BigbearZzz No, since it's required that the $varphi_n$ are $mu$-integrable (the integral of the absolute value is not $infty$)
$endgroup$
– user7802048
Jan 14 at 15:07
$begingroup$
So you're using the word $mu$-integrable in the sense that $int_X |varphi| dmu < infty$? I just want to make sure because I've seen people use it differently.
$endgroup$
– BigbearZzz
Jan 14 at 15:13
$begingroup$
@BigbearZzz Yes exactly. I'll edit my question to include this information.
$endgroup$
– user7802048
Jan 14 at 15:15
add a comment |
$begingroup$
Can $int_X varphi_n dmu = -infty$?
$endgroup$
– BigbearZzz
Jan 14 at 14:38
$begingroup$
@BigbearZzz No, since it's required that the $varphi_n$ are $mu$-integrable (the integral of the absolute value is not $infty$)
$endgroup$
– user7802048
Jan 14 at 15:07
$begingroup$
So you're using the word $mu$-integrable in the sense that $int_X |varphi| dmu < infty$? I just want to make sure because I've seen people use it differently.
$endgroup$
– BigbearZzz
Jan 14 at 15:13
$begingroup$
@BigbearZzz Yes exactly. I'll edit my question to include this information.
$endgroup$
– user7802048
Jan 14 at 15:15
$begingroup$
Can $int_X varphi_n dmu = -infty$?
$endgroup$
– BigbearZzz
Jan 14 at 14:38
$begingroup$
Can $int_X varphi_n dmu = -infty$?
$endgroup$
– BigbearZzz
Jan 14 at 14:38
$begingroup$
@BigbearZzz No, since it's required that the $varphi_n$ are $mu$-integrable (the integral of the absolute value is not $infty$)
$endgroup$
– user7802048
Jan 14 at 15:07
$begingroup$
@BigbearZzz No, since it's required that the $varphi_n$ are $mu$-integrable (the integral of the absolute value is not $infty$)
$endgroup$
– user7802048
Jan 14 at 15:07
$begingroup$
So you're using the word $mu$-integrable in the sense that $int_X |varphi| dmu < infty$? I just want to make sure because I've seen people use it differently.
$endgroup$
– BigbearZzz
Jan 14 at 15:13
$begingroup$
So you're using the word $mu$-integrable in the sense that $int_X |varphi| dmu < infty$? I just want to make sure because I've seen people use it differently.
$endgroup$
– BigbearZzz
Jan 14 at 15:13
$begingroup$
@BigbearZzz Yes exactly. I'll edit my question to include this information.
$endgroup$
– user7802048
Jan 14 at 15:15
$begingroup$
@BigbearZzz Yes exactly. I'll edit my question to include this information.
$endgroup$
– user7802048
Jan 14 at 15:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $varphi_1$ is assumed to be $mu$-integrable, we have
$$
int_X varphi_1^- dmu <infty,
$$
where $f^-(x) := max{0,-f(x)}$ denotes the function's negative part. We can consider instead the sequence
$psi_n:= varphi_n+varphi_1^-$. Obviously, we have $psi_1=varphi_1^+ge 0$ and since $varphi_n$ is increasing, $psi_nge0$ for all $ninBbb N$.
Since $varphi_n to varphi$, we have $psi_nto psi = varphi + varphi_1^-$. By the usual monotone convergence theorem, we can deduce
$$
int_X psi, dmu = lim_{ntoinfty} int_X psi_n ,dmu
$$
which means that
$$
int_X varphi, dmu + int_X varphi_1^-, dmu = lim_{ntoinfty} int_X varphi_n + varphi_1^-,dmu
$$
can we can subtract $int_X varphi_1^- ,dmu$ from both sides to get
$$
int_X varphi, dmu = lim_{ntoinfty} int_X varphi_n ,dmu.
$$
I don't see why you'd need the bound $int_X varphi_n dmu<M$ at all.
answered Jan 14 at 15:49
BigbearZzzBigbearZzz
8,73621652
$endgroup$
$begingroup$
Thank you for the detailed answer!
$endgroup$
– user7802048
Jan 14 at 16:03
add a comment |
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1 Answer
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1 Answer
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active
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oldest
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$begingroup$
Since $varphi_1$ is assumed to be $mu$-integrable, we have
$$
int_X varphi_1^- dmu <infty,
$$
where $f^-(x) := max{0,-f(x)}$ denotes the function's negative part. We can consider instead the sequence
$psi_n:= varphi_n+varphi_1^-$. Obviously, we have $psi_1=varphi_1^+ge 0$ and since $varphi_n$ is increasing, $psi_nge0$ for all $ninBbb N$.
Since $varphi_n to varphi$, we have $psi_nto psi = varphi + varphi_1^-$. By the usual monotone convergence theorem, we can deduce
$$
int_X psi, dmu = lim_{ntoinfty} int_X psi_n ,dmu
$$
which means that
$$
int_X varphi, dmu + int_X varphi_1^-, dmu = lim_{ntoinfty} int_X varphi_n + varphi_1^-,dmu
$$
can we can subtract $int_X varphi_1^- ,dmu$ from both sides to get
$$
int_X varphi, dmu = lim_{ntoinfty} int_X varphi_n ,dmu.
$$
I don't see why you'd need the bound $int_X varphi_n dmu<M$ at all.
answered Jan 14 at 15:49
BigbearZzzBigbearZzz
8,73621652
$endgroup$
$begingroup$
Thank you for the detailed answer!
$endgroup$
– user7802048
Jan 14 at 16:03
add a comment |
$begingroup$
Since $varphi_1$ is assumed to be $mu$-integrable, we have
$$
int_X varphi_1^- dmu <infty,
$$
where $f^-(x) := max{0,-f(x)}$ denotes the function's negative part. We can consider instead the sequence
$psi_n:= varphi_n+varphi_1^-$. Obviously, we have $psi_1=varphi_1^+ge 0$ and since $varphi_n$ is increasing, $psi_nge0$ for all $ninBbb N$.
Since $varphi_n to varphi$, we have $psi_nto psi = varphi + varphi_1^-$. By the usual monotone convergence theorem, we can deduce
$$
int_X psi, dmu = lim_{ntoinfty} int_X psi_n ,dmu
$$
which means that
$$
int_X varphi, dmu + int_X varphi_1^-, dmu = lim_{ntoinfty} int_X varphi_n + varphi_1^-,dmu
$$
can we can subtract $int_X varphi_1^- ,dmu$ from both sides to get
$$
int_X varphi, dmu = lim_{ntoinfty} int_X varphi_n ,dmu.
$$
I don't see why you'd need the bound $int_X varphi_n dmu<M$ at all.
answered Jan 14 at 15:49
BigbearZzzBigbearZzz
8,73621652
$endgroup$
$begingroup$
Thank you for the detailed answer!
$endgroup$
– user7802048
Jan 14 at 16:03
add a comment |
$begingroup$
Since $varphi_1$ is assumed to be $mu$-integrable, we have
$$
int_X varphi_1^- dmu <infty,
$$
where $f^-(x) := max{0,-f(x)}$ denotes the function's negative part. We can consider instead the sequence
$psi_n:= varphi_n+varphi_1^-$. Obviously, we have $psi_1=varphi_1^+ge 0$ and since $varphi_n$ is increasing, $psi_nge0$ for all $ninBbb N$.
Since $varphi_n to varphi$, we have $psi_nto psi = varphi + varphi_1^-$. By the usual monotone convergence theorem, we can deduce
$$
int_X psi, dmu = lim_{ntoinfty} int_X psi_n ,dmu
$$
which means that
$$
int_X varphi, dmu + int_X varphi_1^-, dmu = lim_{ntoinfty} int_X varphi_n + varphi_1^-,dmu
$$
can we can subtract $int_X varphi_1^- ,dmu$ from both sides to get
$$
int_X varphi, dmu = lim_{ntoinfty} int_X varphi_n ,dmu.
$$
I don't see why you'd need the bound $int_X varphi_n dmu<M$ at all.
answered Jan 14 at 15:49
BigbearZzzBigbearZzz
8,73621652
$endgroup$
Since $varphi_1$ is assumed to be $mu$-integrable, we have
$$
int_X varphi_1^- dmu <infty,
$$
where $f^-(x) := max{0,-f(x)}$ denotes the function's negative part. We can consider instead the sequence
$psi_n:= varphi_n+varphi_1^-$. Obviously, we have $psi_1=varphi_1^+ge 0$ and since $varphi_n$ is increasing, $psi_nge0$ for all $ninBbb N$.
Since $varphi_n to varphi$, we have $psi_nto psi = varphi + varphi_1^-$. By the usual monotone convergence theorem, we can deduce
$$
int_X psi, dmu = lim_{ntoinfty} int_X psi_n ,dmu
$$
which means that
$$
int_X varphi, dmu + int_X varphi_1^-, dmu = lim_{ntoinfty} int_X varphi_n + varphi_1^-,dmu
$$
can we can subtract $int_X varphi_1^- ,dmu$ from both sides to get
$$
int_X varphi, dmu = lim_{ntoinfty} int_X varphi_n ,dmu.
$$
I don't see why you'd need the bound $int_X varphi_n dmu<M$ at all.
answered Jan 14 at 15:49
BigbearZzzBigbearZzz
8,73621652
edited Jan 15 at 7:04
answered Jan 14 at 15:49
BigbearZzzBigbearZzz
8,73621652
answered Jan 14 at 15:49
BigbearZzzBigbearZzz
8,73621652
answered Jan 14 at 15:49
BigbearZzzBigbearZzz
8,73621652
8,73621652
$begingroup$
Thank you for the detailed answer!
$endgroup$
– user7802048
Jan 14 at 16:03
add a comment |
$begingroup$
Thank you for the detailed answer!
$endgroup$
– user7802048
Jan 14 at 16:03
$begingroup$
Thank you for the detailed answer!
$endgroup$
– user7802048
Jan 14 at 16:03
$begingroup$
Thank you for the detailed answer!
$endgroup$
– user7802048
Jan 14 at 16:03
add a comment |
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$begingroup$
Can $int_X varphi_n dmu = -infty$?
$endgroup$
– BigbearZzz
Jan 14 at 14:38
$begingroup$
@BigbearZzz No, since it's required that the $varphi_n$ are $mu$-integrable (the integral of the absolute value is not $infty$)
$endgroup$
– user7802048
Jan 14 at 15:07
$begingroup$
So you're using the word $mu$-integrable in the sense that $int_X |varphi| dmu < infty$? I just want to make sure because I've seen people use it differently.
$endgroup$
– BigbearZzz
Jan 14 at 15:13
$begingroup$
@BigbearZzz Yes exactly. I'll edit my question to include this information.
$endgroup$
– user7802048
Jan 14 at 15:15