Mixing
$L^p$ Dominated Convergence Theorem
$begingroup$
I want to prove the $L^p$ Dominated Convergence Theorem which says :
Let ${ f_n }$ be a sequence of measurable functions that converges pointwise a.e. on E to $f$. For$ 1 leq p < infty$, suppose that there is a function $g$ in $L^p(E)$ such that for all $n$, $|f_n|<g$ a.e. on $E$. Prove that ${ f_n } to f$ in $L^p(E)$.
This is my attempt :
$$ |f_n-f|<|g|+|f| since |f_n - f|leq |f_n| + |f| leq |f|+|g| leq 2 max{|f|,|g|} Rightarrow |f_n - f|^p<2^p(|g|^p + |f|^p) $$
If I show that $2^p(|g|^p + |f|^p)$ is integrable over $E$ then the result follows by Lebesgue's Dominated Convergence theorem. But this is where I am having a question : If ${ f_n }$ is a sequence in $L^p(E)$ and ${ f_n } to f$ pointwise a.e. on $E$, does this necessarily mean that $f in L^p(E) $ ? If yes, the result follows after using $2^p(|g|^p + |f|^p)$ as the dominating function in LDC..
ADDED LATER
For the bold-faced question, I tried like this $$|f|=|f-f_n+f_n|leq |f-f_n|+|f_n| Rightarrow int_E |f|^p leq int_E|f-f_n|^p+int_E|f_n|^p$$ Is there a theorem that says if ${f_n}→f$ pointwise a.e. on E, ${f_n}→f$ uniformly on $Esetminus E_0$ where $m(E_0)$ ? If thats true, for any $epsilon$ and for some $N$ , we can say $|f-f_N|^p < epsilon^{p}$ and since $f_n in L^p(E)$, this ensures that the $int_E |f|^p < infty$.
I'd appreciate if you tell me what is going wrong in my statement. Thank you
real-analysis lp-spaces
asked Jan 1 '14 at 2:10
the8thonethe8thone
1,838935
$endgroup$
add a comment |
$begingroup$
I want to prove the $L^p$ Dominated Convergence Theorem which says :
Let ${ f_n }$ be a sequence of measurable functions that converges pointwise a.e. on E to $f$. For$ 1 leq p < infty$, suppose that there is a function $g$ in $L^p(E)$ such that for all $n$, $|f_n|<g$ a.e. on $E$. Prove that ${ f_n } to f$ in $L^p(E)$.
This is my attempt :
$$ |f_n-f|<|g|+|f| since |f_n - f|leq |f_n| + |f| leq |f|+|g| leq 2 max{|f|,|g|} Rightarrow |f_n - f|^p<2^p(|g|^p + |f|^p) $$
If I show that $2^p(|g|^p + |f|^p)$ is integrable over $E$ then the result follows by Lebesgue's Dominated Convergence theorem. But this is where I am having a question : If ${ f_n }$ is a sequence in $L^p(E)$ and ${ f_n } to f$ pointwise a.e. on $E$, does this necessarily mean that $f in L^p(E) $ ? If yes, the result follows after using $2^p(|g|^p + |f|^p)$ as the dominating function in LDC..
ADDED LATER
For the bold-faced question, I tried like this $$|f|=|f-f_n+f_n|leq |f-f_n|+|f_n| Rightarrow int_E |f|^p leq int_E|f-f_n|^p+int_E|f_n|^p$$ Is there a theorem that says if ${f_n}→f$ pointwise a.e. on E, ${f_n}→f$ uniformly on $Esetminus E_0$ where $m(E_0)$ ? If thats true, for any $epsilon$ and for some $N$ , we can say $|f-f_N|^p < epsilon^{p}$ and since $f_n in L^p(E)$, this ensures that the $int_E |f|^p < infty$.
I'd appreciate if you tell me what is going wrong in my statement. Thank you
real-analysis lp-spaces
asked Jan 1 '14 at 2:10
the8thonethe8thone
1,838935
$endgroup$
$begingroup$
$|f_n| < g$ does not imply $|f_n - f| < |g - f|$.
$endgroup$
– Michael Albanese
Jan 1 '14 at 2:17
2
$begingroup$
$|f_n|<g$ implies $|f| leq g$. $|f_n-f|^p<2^p g^p$ so we can apply LDCT.
$endgroup$
– PVAL-inactive
Jan 1 '14 at 3:16
$begingroup$
That's true...I made that too complicated.Thanks.
$endgroup$
– the8thone
Jan 1 '14 at 3:23
add a comment |
$begingroup$
I want to prove the $L^p$ Dominated Convergence Theorem which says :
Let ${ f_n }$ be a sequence of measurable functions that converges pointwise a.e. on E to $f$. For$ 1 leq p < infty$, suppose that there is a function $g$ in $L^p(E)$ such that for all $n$, $|f_n|<g$ a.e. on $E$. Prove that ${ f_n } to f$ in $L^p(E)$.
This is my attempt :
$$ |f_n-f|<|g|+|f| since |f_n - f|leq |f_n| + |f| leq |f|+|g| leq 2 max{|f|,|g|} Rightarrow |f_n - f|^p<2^p(|g|^p + |f|^p) $$
If I show that $2^p(|g|^p + |f|^p)$ is integrable over $E$ then the result follows by Lebesgue's Dominated Convergence theorem. But this is where I am having a question : If ${ f_n }$ is a sequence in $L^p(E)$ and ${ f_n } to f$ pointwise a.e. on $E$, does this necessarily mean that $f in L^p(E) $ ? If yes, the result follows after using $2^p(|g|^p + |f|^p)$ as the dominating function in LDC..
ADDED LATER
For the bold-faced question, I tried like this $$|f|=|f-f_n+f_n|leq |f-f_n|+|f_n| Rightarrow int_E |f|^p leq int_E|f-f_n|^p+int_E|f_n|^p$$ Is there a theorem that says if ${f_n}→f$ pointwise a.e. on E, ${f_n}→f$ uniformly on $Esetminus E_0$ where $m(E_0)$ ? If thats true, for any $epsilon$ and for some $N$ , we can say $|f-f_N|^p < epsilon^{p}$ and since $f_n in L^p(E)$, this ensures that the $int_E |f|^p < infty$.
I'd appreciate if you tell me what is going wrong in my statement. Thank you
real-analysis lp-spaces
asked Jan 1 '14 at 2:10
the8thonethe8thone
1,838935
$endgroup$
I want to prove the $L^p$ Dominated Convergence Theorem which says :
Let ${ f_n }$ be a sequence of measurable functions that converges pointwise a.e. on E to $f$. For$ 1 leq p < infty$, suppose that there is a function $g$ in $L^p(E)$ such that for all $n$, $|f_n|<g$ a.e. on $E$. Prove that ${ f_n } to f$ in $L^p(E)$.
This is my attempt :
$$ |f_n-f|<|g|+|f| since |f_n - f|leq |f_n| + |f| leq |f|+|g| leq 2 max{|f|,|g|} Rightarrow |f_n - f|^p<2^p(|g|^p + |f|^p) $$
If I show that $2^p(|g|^p + |f|^p)$ is integrable over $E$ then the result follows by Lebesgue's Dominated Convergence theorem. But this is where I am having a question : If ${ f_n }$ is a sequence in $L^p(E)$ and ${ f_n } to f$ pointwise a.e. on $E$, does this necessarily mean that $f in L^p(E) $ ? If yes, the result follows after using $2^p(|g|^p + |f|^p)$ as the dominating function in LDC..
ADDED LATER
For the bold-faced question, I tried like this $$|f|=|f-f_n+f_n|leq |f-f_n|+|f_n| Rightarrow int_E |f|^p leq int_E|f-f_n|^p+int_E|f_n|^p$$ Is there a theorem that says if ${f_n}→f$ pointwise a.e. on E, ${f_n}→f$ uniformly on $Esetminus E_0$ where $m(E_0)$ ? If thats true, for any $epsilon$ and for some $N$ , we can say $|f-f_N|^p < epsilon^{p}$ and since $f_n in L^p(E)$, this ensures that the $int_E |f|^p < infty$.
I'd appreciate if you tell me what is going wrong in my statement. Thank you
real-analysis lp-spaces
real-analysis lp-spaces
asked Jan 1 '14 at 2:10
the8thonethe8thone
1,838935
asked Jan 1 '14 at 2:10
the8thonethe8thone
1,838935
edited Jan 1 '14 at 2:52
the8thone
asked Jan 1 '14 at 2:10
the8thonethe8thone
1,838935
asked Jan 1 '14 at 2:10
the8thonethe8thone
1,838935
asked Jan 1 '14 at 2:10
the8thonethe8thone
1,838935
1,838935
$begingroup$
$|f_n| < g$ does not imply $|f_n - f| < |g - f|$.
$endgroup$
– Michael Albanese
Jan 1 '14 at 2:17
2
$begingroup$
$|f_n|<g$ implies $|f| leq g$. $|f_n-f|^p<2^p g^p$ so we can apply LDCT.
$endgroup$
– PVAL-inactive
Jan 1 '14 at 3:16
$begingroup$
That's true...I made that too complicated.Thanks.
$endgroup$
– the8thone
Jan 1 '14 at 3:23
add a comment |
$begingroup$
$|f_n| < g$ does not imply $|f_n - f| < |g - f|$.
$endgroup$
– Michael Albanese
Jan 1 '14 at 2:17
2
$begingroup$
$|f_n|<g$ implies $|f| leq g$. $|f_n-f|^p<2^p g^p$ so we can apply LDCT.
$endgroup$
– PVAL-inactive
Jan 1 '14 at 3:16
$begingroup$
That's true...I made that too complicated.Thanks.
$endgroup$
– the8thone
Jan 1 '14 at 3:23
$begingroup$
$|f_n| < g$ does not imply $|f_n - f| < |g - f|$.
$endgroup$
– Michael Albanese
Jan 1 '14 at 2:17
$begingroup$
$|f_n| < g$ does not imply $|f_n - f| < |g - f|$.
$endgroup$
– Michael Albanese
Jan 1 '14 at 2:17
2
2
$begingroup$
$|f_n|<g$ implies $|f| leq g$. $|f_n-f|^p<2^p g^p$ so we can apply LDCT.
$endgroup$
– PVAL-inactive
Jan 1 '14 at 3:16
$begingroup$
$|f_n|<g$ implies $|f| leq g$. $|f_n-f|^p<2^p g^p$ so we can apply LDCT.
$endgroup$
– PVAL-inactive
Jan 1 '14 at 3:16
$begingroup$
That's true...I made that too complicated.Thanks.
$endgroup$
– the8thone
Jan 1 '14 at 3:23
$begingroup$
That's true...I made that too complicated.Thanks.
$endgroup$
– the8thone
Jan 1 '14 at 3:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For your additional question, we have Egorov (sometimes spelled Egoroff) theorem:
On a probability measure space $(Omega, {cal F}, mu)$, if $(X_n)_{ngeq1}$ is a sequence of random variables convergent to a random variable $X$ $mu$-a.e, then for any given $delta>0$ there is an event $Ain {cal F}$ such that $mu(A) < delta$ and $(X_n)_{ngeq1}$ converges uniformly to $X$ on $Omega setminus A$.
Best regards
answered Jan 1 '14 at 4:54
ir7ir7
4,19811115
$endgroup$
add a comment |
$begingroup$
I think I have simpler proof.
If we know that $ f_n to f $ in measure this means that there is subsequence $f_{n_k}$ converging to $f$ point wise a.e. We still have $left| f_{n_k} right| < g $ so the Dominated Convergence (for $L_1$) implies that
$$
intop left| f_{n_k} - f right|dmu to 0
$$
We know (form the point-wise converges) that $left| f_{n_k} - f right| to 0 $ a.e so we can assume that $left| f_{n_k} - f right| < 1 $ a.e. This implies us
$$
0leq intop left| f_{n_k} - f right| ^ p dmu leq intop left| f_{n_k} - f right|dmu to 0
$$ Thus
$$
leftVert f_{n_k} -f rightVert _p to 0
$$
If we won't have $ leftVert f_{n} -f rightVert _p to 0 $ (i.e. converges in $L^p$) we could construct subsequence of $f_n$ $h_i = f_{n_i}$ such that
$$
leftVert h_{i} -f rightVert _p > epsilon
$$
for some $epsilon > 0 $. We would still have $leftvert h_i rightvert < g$ and $h_i to f$ is measure, so we may construct subsequence $h_{i_k}$ (as done before) of $h_i$ such that
$$
leftVert h_{i_k} - f rightVert _p to 0
$$
which contradict the condition that defined $h_i$.
answered Jan 30 at 22:45
Barak OhanaBarak Ohana
225
$endgroup$
add a comment |
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2 Answers
2
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oldest
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
For your additional question, we have Egorov (sometimes spelled Egoroff) theorem:
On a probability measure space $(Omega, {cal F}, mu)$, if $(X_n)_{ngeq1}$ is a sequence of random variables convergent to a random variable $X$ $mu$-a.e, then for any given $delta>0$ there is an event $Ain {cal F}$ such that $mu(A) < delta$ and $(X_n)_{ngeq1}$ converges uniformly to $X$ on $Omega setminus A$.
Best regards
answered Jan 1 '14 at 4:54
ir7ir7
4,19811115
$endgroup$
add a comment |
$begingroup$
For your additional question, we have Egorov (sometimes spelled Egoroff) theorem:
On a probability measure space $(Omega, {cal F}, mu)$, if $(X_n)_{ngeq1}$ is a sequence of random variables convergent to a random variable $X$ $mu$-a.e, then for any given $delta>0$ there is an event $Ain {cal F}$ such that $mu(A) < delta$ and $(X_n)_{ngeq1}$ converges uniformly to $X$ on $Omega setminus A$.
Best regards
answered Jan 1 '14 at 4:54
ir7ir7
4,19811115
$endgroup$
add a comment |
$begingroup$
For your additional question, we have Egorov (sometimes spelled Egoroff) theorem:
On a probability measure space $(Omega, {cal F}, mu)$, if $(X_n)_{ngeq1}$ is a sequence of random variables convergent to a random variable $X$ $mu$-a.e, then for any given $delta>0$ there is an event $Ain {cal F}$ such that $mu(A) < delta$ and $(X_n)_{ngeq1}$ converges uniformly to $X$ on $Omega setminus A$.
Best regards
answered Jan 1 '14 at 4:54
ir7ir7
4,19811115
$endgroup$
For your additional question, we have Egorov (sometimes spelled Egoroff) theorem:
On a probability measure space $(Omega, {cal F}, mu)$, if $(X_n)_{ngeq1}$ is a sequence of random variables convergent to a random variable $X$ $mu$-a.e, then for any given $delta>0$ there is an event $Ain {cal F}$ such that $mu(A) < delta$ and $(X_n)_{ngeq1}$ converges uniformly to $X$ on $Omega setminus A$.
Best regards
answered Jan 1 '14 at 4:54
ir7ir7
4,19811115
answered Jan 1 '14 at 4:54
ir7ir7
4,19811115
answered Jan 1 '14 at 4:54
ir7ir7
4,19811115
answered Jan 1 '14 at 4:54
ir7ir7
4,19811115
4,19811115
add a comment |
add a comment |
$begingroup$
I think I have simpler proof.
If we know that $ f_n to f $ in measure this means that there is subsequence $f_{n_k}$ converging to $f$ point wise a.e. We still have $left| f_{n_k} right| < g $ so the Dominated Convergence (for $L_1$) implies that
$$
intop left| f_{n_k} - f right|dmu to 0
$$
We know (form the point-wise converges) that $left| f_{n_k} - f right| to 0 $ a.e so we can assume that $left| f_{n_k} - f right| < 1 $ a.e. This implies us
$$
0leq intop left| f_{n_k} - f right| ^ p dmu leq intop left| f_{n_k} - f right|dmu to 0
$$ Thus
$$
leftVert f_{n_k} -f rightVert _p to 0
$$
If we won't have $ leftVert f_{n} -f rightVert _p to 0 $ (i.e. converges in $L^p$) we could construct subsequence of $f_n$ $h_i = f_{n_i}$ such that
$$
leftVert h_{i} -f rightVert _p > epsilon
$$
for some $epsilon > 0 $. We would still have $leftvert h_i rightvert < g$ and $h_i to f$ is measure, so we may construct subsequence $h_{i_k}$ (as done before) of $h_i$ such that
$$
leftVert h_{i_k} - f rightVert _p to 0
$$
which contradict the condition that defined $h_i$.
answered Jan 30 at 22:45
Barak OhanaBarak Ohana
225
$endgroup$
add a comment |
$begingroup$
I think I have simpler proof.
If we know that $ f_n to f $ in measure this means that there is subsequence $f_{n_k}$ converging to $f$ point wise a.e. We still have $left| f_{n_k} right| < g $ so the Dominated Convergence (for $L_1$) implies that
$$
intop left| f_{n_k} - f right|dmu to 0
$$
We know (form the point-wise converges) that $left| f_{n_k} - f right| to 0 $ a.e so we can assume that $left| f_{n_k} - f right| < 1 $ a.e. This implies us
$$
0leq intop left| f_{n_k} - f right| ^ p dmu leq intop left| f_{n_k} - f right|dmu to 0
$$ Thus
$$
leftVert f_{n_k} -f rightVert _p to 0
$$
If we won't have $ leftVert f_{n} -f rightVert _p to 0 $ (i.e. converges in $L^p$) we could construct subsequence of $f_n$ $h_i = f_{n_i}$ such that
$$
leftVert h_{i} -f rightVert _p > epsilon
$$
for some $epsilon > 0 $. We would still have $leftvert h_i rightvert < g$ and $h_i to f$ is measure, so we may construct subsequence $h_{i_k}$ (as done before) of $h_i$ such that
$$
leftVert h_{i_k} - f rightVert _p to 0
$$
which contradict the condition that defined $h_i$.
answered Jan 30 at 22:45
Barak OhanaBarak Ohana
225
$endgroup$
add a comment |
$begingroup$
I think I have simpler proof.
If we know that $ f_n to f $ in measure this means that there is subsequence $f_{n_k}$ converging to $f$ point wise a.e. We still have $left| f_{n_k} right| < g $ so the Dominated Convergence (for $L_1$) implies that
$$
intop left| f_{n_k} - f right|dmu to 0
$$
We know (form the point-wise converges) that $left| f_{n_k} - f right| to 0 $ a.e so we can assume that $left| f_{n_k} - f right| < 1 $ a.e. This implies us
$$
0leq intop left| f_{n_k} - f right| ^ p dmu leq intop left| f_{n_k} - f right|dmu to 0
$$ Thus
$$
leftVert f_{n_k} -f rightVert _p to 0
$$
If we won't have $ leftVert f_{n} -f rightVert _p to 0 $ (i.e. converges in $L^p$) we could construct subsequence of $f_n$ $h_i = f_{n_i}$ such that
$$
leftVert h_{i} -f rightVert _p > epsilon
$$
for some $epsilon > 0 $. We would still have $leftvert h_i rightvert < g$ and $h_i to f$ is measure, so we may construct subsequence $h_{i_k}$ (as done before) of $h_i$ such that
$$
leftVert h_{i_k} - f rightVert _p to 0
$$
which contradict the condition that defined $h_i$.
answered Jan 30 at 22:45
Barak OhanaBarak Ohana
225
$endgroup$
I think I have simpler proof.
If we know that $ f_n to f $ in measure this means that there is subsequence $f_{n_k}$ converging to $f$ point wise a.e. We still have $left| f_{n_k} right| < g $ so the Dominated Convergence (for $L_1$) implies that
$$
intop left| f_{n_k} - f right|dmu to 0
$$
We know (form the point-wise converges) that $left| f_{n_k} - f right| to 0 $ a.e so we can assume that $left| f_{n_k} - f right| < 1 $ a.e. This implies us
$$
0leq intop left| f_{n_k} - f right| ^ p dmu leq intop left| f_{n_k} - f right|dmu to 0
$$ Thus
$$
leftVert f_{n_k} -f rightVert _p to 0
$$
If we won't have $ leftVert f_{n} -f rightVert _p to 0 $ (i.e. converges in $L^p$) we could construct subsequence of $f_n$ $h_i = f_{n_i}$ such that
$$
leftVert h_{i} -f rightVert _p > epsilon
$$
for some $epsilon > 0 $. We would still have $leftvert h_i rightvert < g$ and $h_i to f$ is measure, so we may construct subsequence $h_{i_k}$ (as done before) of $h_i$ such that
$$
leftVert h_{i_k} - f rightVert _p to 0
$$
which contradict the condition that defined $h_i$.
answered Jan 30 at 22:45
Barak OhanaBarak Ohana
225
edited Jan 31 at 17:33
answered Jan 30 at 22:45
Barak OhanaBarak Ohana
225
answered Jan 30 at 22:45
Barak OhanaBarak Ohana
225
answered Jan 30 at 22:45
Barak OhanaBarak Ohana
225
225
add a comment |
add a comment |
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$begingroup$
$|f_n| < g$ does not imply $|f_n - f| < |g - f|$.
$endgroup$
– Michael Albanese
Jan 1 '14 at 2:17
2
$begingroup$
$|f_n|<g$ implies $|f| leq g$. $|f_n-f|^p<2^p g^p$ so we can apply LDCT.
$endgroup$
– PVAL-inactive
Jan 1 '14 at 3:16
$begingroup$
That's true...I made that too complicated.Thanks.
$endgroup$
– the8thone
Jan 1 '14 at 3:23