Does $f_n to f$ pointwisely imply $int f_n to int f$ for conditionally Riemann integrable functions?
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Let $f_n:mathbb{R}^ntomathbb{R}$ be a sequence of conditionally (improper) Riemann integrable functions that pointwisely converges to $f:mathbb{R}^ntomathbb{R}$ which is also conditionally Riemann integrable. Does $int f_n to int f$? If they were absolutely Riemann integrable or Lebesgue integrable, it would be true by the dominated convergence theorem. I'd like to know it is also true (with some additional conditions) for conditionally converging integrals.
calculus real-analysis
asked yesterday
rimusolem
564
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up vote
1
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Let $f_n:mathbb{R}^ntomathbb{R}$ be a sequence of conditionally (improper) Riemann integrable functions that pointwisely converges to $f:mathbb{R}^ntomathbb{R}$ which is also conditionally Riemann integrable. Does $int f_n to int f$? If they were absolutely Riemann integrable or Lebesgue integrable, it would be true by the dominated convergence theorem. I'd like to know it is also true (with some additional conditions) for conditionally converging integrals.
calculus real-analysis
asked yesterday
rimusolem
564
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f_n:mathbb{R}^ntomathbb{R}$ be a sequence of conditionally (improper) Riemann integrable functions that pointwisely converges to $f:mathbb{R}^ntomathbb{R}$ which is also conditionally Riemann integrable. Does $int f_n to int f$? If they were absolutely Riemann integrable or Lebesgue integrable, it would be true by the dominated convergence theorem. I'd like to know it is also true (with some additional conditions) for conditionally converging integrals.
calculus real-analysis
asked yesterday
rimusolem
564
Let $f_n:mathbb{R}^ntomathbb{R}$ be a sequence of conditionally (improper) Riemann integrable functions that pointwisely converges to $f:mathbb{R}^ntomathbb{R}$ which is also conditionally Riemann integrable. Does $int f_n to int f$? If they were absolutely Riemann integrable or Lebesgue integrable, it would be true by the dominated convergence theorem. I'd like to know it is also true (with some additional conditions) for conditionally converging integrals.
calculus real-analysis
calculus real-analysis
asked yesterday
rimusolem
564
asked yesterday
rimusolem
564
asked yesterday
rimusolem
564
asked yesterday
rimusolem
564
asked yesterday
rimusolem
564
564
add a comment |
add a comment |
1 Answer
1
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2
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accepted
It's not true, even for absolutely Riemann integrable functions as you stated.
Take $$f_n = n1_{left(0,frac{1}{n}right]}$$
then $f_n to f$ pointwise with $$f equiv 0$$
But $$int f_n = 1 notto 0 = int f$$
You cannot use dominated convergence theorem here although absolutely riemann integrability… why?
answered yesterday
Gono
3,524416
How about $0<int |f| <infty$? Can this be a sufficient condition? Intuitively, this may prohibit accumulation of the mass of $f_n$ in an infinitely small region.
– rimusolem
yesterday
I opened a spearate question for the above comment.
– rimusolem
yesterday
OFC not… take $$tilde{f_n} = 1_{[0,1]} + f_n$$ then $$ tilde{f} = 1_{[0,1]}$$ and so your condition holds but still $$int tilde{f_n} = 2 notto 1 = int tilde{f}$$
– Gono
21 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It's not true, even for absolutely Riemann integrable functions as you stated.
Take $$f_n = n1_{left(0,frac{1}{n}right]}$$
then $f_n to f$ pointwise with $$f equiv 0$$
But $$int f_n = 1 notto 0 = int f$$
You cannot use dominated convergence theorem here although absolutely riemann integrability… why?
answered yesterday
Gono
3,524416
How about $0<int |f| <infty$? Can this be a sufficient condition? Intuitively, this may prohibit accumulation of the mass of $f_n$ in an infinitely small region.
– rimusolem
yesterday
I opened a spearate question for the above comment.
– rimusolem
yesterday
OFC not… take $$tilde{f_n} = 1_{[0,1]} + f_n$$ then $$ tilde{f} = 1_{[0,1]}$$ and so your condition holds but still $$int tilde{f_n} = 2 notto 1 = int tilde{f}$$
– Gono
21 hours ago
add a comment |
up vote
2
down vote
accepted
It's not true, even for absolutely Riemann integrable functions as you stated.
Take $$f_n = n1_{left(0,frac{1}{n}right]}$$
then $f_n to f$ pointwise with $$f equiv 0$$
But $$int f_n = 1 notto 0 = int f$$
You cannot use dominated convergence theorem here although absolutely riemann integrability… why?
answered yesterday
Gono
3,524416
How about $0<int |f| <infty$? Can this be a sufficient condition? Intuitively, this may prohibit accumulation of the mass of $f_n$ in an infinitely small region.
– rimusolem
yesterday
I opened a spearate question for the above comment.
– rimusolem
yesterday
OFC not… take $$tilde{f_n} = 1_{[0,1]} + f_n$$ then $$ tilde{f} = 1_{[0,1]}$$ and so your condition holds but still $$int tilde{f_n} = 2 notto 1 = int tilde{f}$$
– Gono
21 hours ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It's not true, even for absolutely Riemann integrable functions as you stated.
Take $$f_n = n1_{left(0,frac{1}{n}right]}$$
then $f_n to f$ pointwise with $$f equiv 0$$
But $$int f_n = 1 notto 0 = int f$$
You cannot use dominated convergence theorem here although absolutely riemann integrability… why?
answered yesterday
Gono
3,524416
It's not true, even for absolutely Riemann integrable functions as you stated.
Take $$f_n = n1_{left(0,frac{1}{n}right]}$$
then $f_n to f$ pointwise with $$f equiv 0$$
But $$int f_n = 1 notto 0 = int f$$
You cannot use dominated convergence theorem here although absolutely riemann integrability… why?
answered yesterday
Gono
3,524416
answered yesterday
Gono
3,524416
answered yesterday
Gono
3,524416
answered yesterday
Gono
3,524416
3,524416
How about $0<int |f| <infty$? Can this be a sufficient condition? Intuitively, this may prohibit accumulation of the mass of $f_n$ in an infinitely small region.
– rimusolem
yesterday
I opened a spearate question for the above comment.
– rimusolem
yesterday
OFC not… take $$tilde{f_n} = 1_{[0,1]} + f_n$$ then $$ tilde{f} = 1_{[0,1]}$$ and so your condition holds but still $$int tilde{f_n} = 2 notto 1 = int tilde{f}$$
– Gono
21 hours ago
add a comment |
How about $0<int |f| <infty$? Can this be a sufficient condition? Intuitively, this may prohibit accumulation of the mass of $f_n$ in an infinitely small region.
– rimusolem
yesterday
I opened a spearate question for the above comment.
– rimusolem
yesterday
OFC not… take $$tilde{f_n} = 1_{[0,1]} + f_n$$ then $$ tilde{f} = 1_{[0,1]}$$ and so your condition holds but still $$int tilde{f_n} = 2 notto 1 = int tilde{f}$$
– Gono
21 hours ago
How about $0<int |f| <infty$? Can this be a sufficient condition? Intuitively, this may prohibit accumulation of the mass of $f_n$ in an infinitely small region.
– rimusolem
yesterday
How about $0<int |f| <infty$? Can this be a sufficient condition? Intuitively, this may prohibit accumulation of the mass of $f_n$ in an infinitely small region.
– rimusolem
yesterday
I opened a spearate question for the above comment.
– rimusolem
yesterday
I opened a spearate question for the above comment.
– rimusolem
yesterday
OFC not… take $$tilde{f_n} = 1_{[0,1]} + f_n$$ then $$ tilde{f} = 1_{[0,1]}$$ and so your condition holds but still $$int tilde{f_n} = 2 notto 1 = int tilde{f}$$
– Gono
21 hours ago
OFC not… take $$tilde{f_n} = 1_{[0,1]} + f_n$$ then $$ tilde{f} = 1_{[0,1]}$$ and so your condition holds but still $$int tilde{f_n} = 2 notto 1 = int tilde{f}$$
– Gono
21 hours ago
add a comment |
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