Simple question about convergence a.e
How could following property be proved without using Dominated Convergence Theorem?
Let $left{{f_n(x)}right}$ ($f_n:mathbb{R^m}rightarrow{}mathbb{R}$) be a sequence of functions such that $f_nrightarrow{f};a.e$ and $|f_n|leq{g}$ with $g$ an integrable function.
Then $f_nrightarrow{f}$ in $L_1(mathbb{R^n})$.
Does someone know a counterexample if it isn't true the second condition?
Thanks.
real-analysis functional-analysis analysis
edited Dec 29 '18 at 10:43
mathcounterexamples.net
25.3k21953
asked Dec 29 '18 at 10:32
mathlifemathlife
629
add a comment |
How could following property be proved without using Dominated Convergence Theorem?
Let $left{{f_n(x)}right}$ ($f_n:mathbb{R^m}rightarrow{}mathbb{R}$) be a sequence of functions such that $f_nrightarrow{f};a.e$ and $|f_n|leq{g}$ with $g$ an integrable function.
Then $f_nrightarrow{f}$ in $L_1(mathbb{R^n})$.
Does someone know a counterexample if it isn't true the second condition?
Thanks.
real-analysis functional-analysis analysis
edited Dec 29 '18 at 10:43
mathcounterexamples.net
25.3k21953
asked Dec 29 '18 at 10:32
mathlifemathlife
629
Your question basically is: Prove the dominated convergence theorem without using the dominated convergence theorem.
– Math_QED
Dec 29 '18 at 10:57
Ok ok. I understand
– mathlife
Dec 29 '18 at 11:02
So for a proof, just look at any basic measure theory book (Rudin's POMA/RCA - Folland's Real analysis - Royden's Real analysis are examples that come to mind).
– Math_QED
Dec 29 '18 at 11:05
add a comment |
How could following property be proved without using Dominated Convergence Theorem?
Let $left{{f_n(x)}right}$ ($f_n:mathbb{R^m}rightarrow{}mathbb{R}$) be a sequence of functions such that $f_nrightarrow{f};a.e$ and $|f_n|leq{g}$ with $g$ an integrable function.
Then $f_nrightarrow{f}$ in $L_1(mathbb{R^n})$.
Does someone know a counterexample if it isn't true the second condition?
Thanks.
real-analysis functional-analysis analysis
edited Dec 29 '18 at 10:43
mathcounterexamples.net
25.3k21953
asked Dec 29 '18 at 10:32
mathlifemathlife
629
How could following property be proved without using Dominated Convergence Theorem?
Let $left{{f_n(x)}right}$ ($f_n:mathbb{R^m}rightarrow{}mathbb{R}$) be a sequence of functions such that $f_nrightarrow{f};a.e$ and $|f_n|leq{g}$ with $g$ an integrable function.
Then $f_nrightarrow{f}$ in $L_1(mathbb{R^n})$.
Does someone know a counterexample if it isn't true the second condition?
Thanks.
real-analysis functional-analysis analysis
real-analysis functional-analysis analysis
edited Dec 29 '18 at 10:43
mathcounterexamples.net
25.3k21953
asked Dec 29 '18 at 10:32
mathlifemathlife
629
edited Dec 29 '18 at 10:43
mathcounterexamples.net
25.3k21953
asked Dec 29 '18 at 10:32
mathlifemathlife
629
edited Dec 29 '18 at 10:43
mathcounterexamples.net
25.3k21953
edited Dec 29 '18 at 10:43
mathcounterexamples.net
25.3k21953
edited Dec 29 '18 at 10:43
mathcounterexamples.net
25.3k21953
25.3k21953
asked Dec 29 '18 at 10:32
mathlifemathlife
629
asked Dec 29 '18 at 10:32
mathlifemathlife
629
asked Dec 29 '18 at 10:32
mathlifemathlife
629
629
Your question basically is: Prove the dominated convergence theorem without using the dominated convergence theorem.
– Math_QED
Dec 29 '18 at 10:57
Ok ok. I understand
– mathlife
Dec 29 '18 at 11:02
So for a proof, just look at any basic measure theory book (Rudin's POMA/RCA - Folland's Real analysis - Royden's Real analysis are examples that come to mind).
– Math_QED
Dec 29 '18 at 11:05
add a comment |
Your question basically is: Prove the dominated convergence theorem without using the dominated convergence theorem.
– Math_QED
Dec 29 '18 at 10:57
Ok ok. I understand
– mathlife
Dec 29 '18 at 11:02
So for a proof, just look at any basic measure theory book (Rudin's POMA/RCA - Folland's Real analysis - Royden's Real analysis are examples that come to mind).
– Math_QED
Dec 29 '18 at 11:05
Your question basically is: Prove the dominated convergence theorem without using the dominated convergence theorem.
– Math_QED
Dec 29 '18 at 10:57
Your question basically is: Prove the dominated convergence theorem without using the dominated convergence theorem.
– Math_QED
Dec 29 '18 at 10:57
Ok ok. I understand
– mathlife
Dec 29 '18 at 11:02
Ok ok. I understand
– mathlife
Dec 29 '18 at 11:02
So for a proof, just look at any basic measure theory book (Rudin's POMA/RCA - Folland's Real analysis - Royden's Real analysis are examples that come to mind).
– Math_QED
Dec 29 '18 at 11:05
So for a proof, just look at any basic measure theory book (Rudin's POMA/RCA - Folland's Real analysis - Royden's Real analysis are examples that come to mind).
– Math_QED
Dec 29 '18 at 11:05
add a comment |
1 Answer
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The property you are stating is the Dominated Convergence theorem: https://en.wikipedia.org/wiki/Dominated_convergence_theorem
A simple counter example is the sequence of functions $f_n: mathbb R to mathbb R$ defined by $$f_n(x) := mathbb 1_{[n,n+1[}.$$
Then $f_n to 0$ pointwise, but $$intlvert f_n - 0 rvert , dlambda = 1 quad forall nin mathbb N.$$
Note that there does not exist an integrable $g$ as in the statement of your property.
answered Dec 29 '18 at 10:46
bavor42bavor42
30419
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
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votes
The property you are stating is the Dominated Convergence theorem: https://en.wikipedia.org/wiki/Dominated_convergence_theorem
A simple counter example is the sequence of functions $f_n: mathbb R to mathbb R$ defined by $$f_n(x) := mathbb 1_{[n,n+1[}.$$
Then $f_n to 0$ pointwise, but $$intlvert f_n - 0 rvert , dlambda = 1 quad forall nin mathbb N.$$
Note that there does not exist an integrable $g$ as in the statement of your property.
answered Dec 29 '18 at 10:46
bavor42bavor42
30419
add a comment |
The property you are stating is the Dominated Convergence theorem: https://en.wikipedia.org/wiki/Dominated_convergence_theorem
A simple counter example is the sequence of functions $f_n: mathbb R to mathbb R$ defined by $$f_n(x) := mathbb 1_{[n,n+1[}.$$
Then $f_n to 0$ pointwise, but $$intlvert f_n - 0 rvert , dlambda = 1 quad forall nin mathbb N.$$
Note that there does not exist an integrable $g$ as in the statement of your property.
answered Dec 29 '18 at 10:46
bavor42bavor42
30419
add a comment |
The property you are stating is the Dominated Convergence theorem: https://en.wikipedia.org/wiki/Dominated_convergence_theorem
A simple counter example is the sequence of functions $f_n: mathbb R to mathbb R$ defined by $$f_n(x) := mathbb 1_{[n,n+1[}.$$
Then $f_n to 0$ pointwise, but $$intlvert f_n - 0 rvert , dlambda = 1 quad forall nin mathbb N.$$
Note that there does not exist an integrable $g$ as in the statement of your property.
answered Dec 29 '18 at 10:46
bavor42bavor42
30419
The property you are stating is the Dominated Convergence theorem: https://en.wikipedia.org/wiki/Dominated_convergence_theorem
A simple counter example is the sequence of functions $f_n: mathbb R to mathbb R$ defined by $$f_n(x) := mathbb 1_{[n,n+1[}.$$
Then $f_n to 0$ pointwise, but $$intlvert f_n - 0 rvert , dlambda = 1 quad forall nin mathbb N.$$
Note that there does not exist an integrable $g$ as in the statement of your property.
answered Dec 29 '18 at 10:46
bavor42bavor42
30419
edited Dec 29 '18 at 10:52
answered Dec 29 '18 at 10:46
bavor42bavor42
30419
answered Dec 29 '18 at 10:46
bavor42bavor42
30419
answered Dec 29 '18 at 10:46
bavor42bavor42
30419
30419
add a comment |
add a comment |
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Your question basically is: Prove the dominated convergence theorem without using the dominated convergence theorem.
– Math_QED
Dec 29 '18 at 10:57
Ok ok. I understand
– mathlife
Dec 29 '18 at 11:02
So for a proof, just look at any basic measure theory book (Rudin's POMA/RCA - Folland's Real analysis - Royden's Real analysis are examples that come to mind).
– Math_QED
Dec 29 '18 at 11:05