Let f be a real-valued function of two variables (x,y) that is defined on the square
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Let $f$ be a real-valued function of two variables $(x,y)$ that is defined on the square $Q={(x,y)| 0 leq x leq 1, 0 leq y leq 1}$ and is a measurable function of x for each fixed value of y. Suppose for each fixed value of x, $lim_{y to 0} f(x,y) = f(x)$ and that for all y, we have $|f(x,y)| leq g(x)$, where g is integrable over [0,1]. Show that
$lim_{y to 0} $$int_{0}^{1} f(x,y) dx$$ = $$int_{0}^{1} f(x) dx$$ $
Also show that if the function f(x,y) is continuous in y for each x, then
$h(y)= $$int_{0}^{1} f(x,y) dx$$ $ is a continuous function of y
measure-theory
asked Nov 15 at 23:48
Sarah2018
191
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Let $f$ be a real-valued function of two variables $(x,y)$ that is defined on the square $Q={(x,y)| 0 leq x leq 1, 0 leq y leq 1}$ and is a measurable function of x for each fixed value of y. Suppose for each fixed value of x, $lim_{y to 0} f(x,y) = f(x)$ and that for all y, we have $|f(x,y)| leq g(x)$, where g is integrable over [0,1]. Show that
$lim_{y to 0} $$int_{0}^{1} f(x,y) dx$$ = $$int_{0}^{1} f(x) dx$$ $
Also show that if the function f(x,y) is continuous in y for each x, then
$h(y)= $$int_{0}^{1} f(x,y) dx$$ $ is a continuous function of y
measure-theory
asked Nov 15 at 23:48
Sarah2018
191
If you can show that $h$ is continuous (not hard), then you have that $h$ is integrable, and then DCT is an easy thing to use.
– Sean Roberson
Nov 16 at 0:29
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up vote
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favorite
Let $f$ be a real-valued function of two variables $(x,y)$ that is defined on the square $Q={(x,y)| 0 leq x leq 1, 0 leq y leq 1}$ and is a measurable function of x for each fixed value of y. Suppose for each fixed value of x, $lim_{y to 0} f(x,y) = f(x)$ and that for all y, we have $|f(x,y)| leq g(x)$, where g is integrable over [0,1]. Show that
$lim_{y to 0} $$int_{0}^{1} f(x,y) dx$$ = $$int_{0}^{1} f(x) dx$$ $
Also show that if the function f(x,y) is continuous in y for each x, then
$h(y)= $$int_{0}^{1} f(x,y) dx$$ $ is a continuous function of y
measure-theory
asked Nov 15 at 23:48
Sarah2018
191
Let $f$ be a real-valued function of two variables $(x,y)$ that is defined on the square $Q={(x,y)| 0 leq x leq 1, 0 leq y leq 1}$ and is a measurable function of x for each fixed value of y. Suppose for each fixed value of x, $lim_{y to 0} f(x,y) = f(x)$ and that for all y, we have $|f(x,y)| leq g(x)$, where g is integrable over [0,1]. Show that
$lim_{y to 0} $$int_{0}^{1} f(x,y) dx$$ = $$int_{0}^{1} f(x) dx$$ $
Also show that if the function f(x,y) is continuous in y for each x, then
$h(y)= $$int_{0}^{1} f(x,y) dx$$ $ is a continuous function of y
measure-theory
measure-theory
asked Nov 15 at 23:48
Sarah2018
191
asked Nov 15 at 23:48
Sarah2018
191
asked Nov 15 at 23:48
Sarah2018
191
asked Nov 15 at 23:48
Sarah2018
191
asked Nov 15 at 23:48
Sarah2018
191
191
If you can show that $h$ is continuous (not hard), then you have that $h$ is integrable, and then DCT is an easy thing to use.
– Sean Roberson
Nov 16 at 0:29
add a comment |
If you can show that $h$ is continuous (not hard), then you have that $h$ is integrable, and then DCT is an easy thing to use.
– Sean Roberson
Nov 16 at 0:29
If you can show that $h$ is continuous (not hard), then you have that $h$ is integrable, and then DCT is an easy thing to use.
– Sean Roberson
Nov 16 at 0:29
If you can show that $h$ is continuous (not hard), then you have that $h$ is integrable, and then DCT is an easy thing to use.
– Sean Roberson
Nov 16 at 0:29
add a comment |
1 Answer
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Do you know Dominated Convergence Theorem? Both parts of your question are immediate applications of this theorem.
answered Nov 16 at 0:26
Kavi Rama Murthy
42k31751
add a comment |
1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Do you know Dominated Convergence Theorem? Both parts of your question are immediate applications of this theorem.
answered Nov 16 at 0:26
Kavi Rama Murthy
42k31751
add a comment |
up vote
2
down vote
Do you know Dominated Convergence Theorem? Both parts of your question are immediate applications of this theorem.
answered Nov 16 at 0:26
Kavi Rama Murthy
42k31751
add a comment |
up vote
2
down vote
up vote
2
down vote
Do you know Dominated Convergence Theorem? Both parts of your question are immediate applications of this theorem.
answered Nov 16 at 0:26
Kavi Rama Murthy
42k31751
Do you know Dominated Convergence Theorem? Both parts of your question are immediate applications of this theorem.
answered Nov 16 at 0:26
Kavi Rama Murthy
42k31751
answered Nov 16 at 0:26
Kavi Rama Murthy
42k31751
answered Nov 16 at 0:26
Kavi Rama Murthy
42k31751
answered Nov 16 at 0:26
Kavi Rama Murthy
42k31751
42k31751
add a comment |
add a comment |
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If you can show that $h$ is continuous (not hard), then you have that $h$ is integrable, and then DCT is an easy thing to use.
– Sean Roberson
Nov 16 at 0:29