Mixing
If $f_n$ $longrightarrow$ $f$ in measure and $g_n$ $longrightarrow$ $g$ in measure then show that $f_n +g_n$...
(1)
If $f_n$ $longrightarrow$ $f$ in measure and $g_n$ $longrightarrow$ $g$ in measure then show that $f_n +g_n$ $longrightarrow$ $f+g$ in measure
(2)If $f_n longrightarrow theta$ in measure and $g_n longrightarrow theta$ in measure where $theta$ is the identically zero function on $Omega$ then show that $f_ng_n longrightarrow 0$
My attempt
For both cases I use the fact that $(Omega,Sigma,mu)$ is my measure space where everything is defined in the H.L Royden.
For (1) it follows from the relation $ {x in E |(f_n+g_n)(x)-(f+g)(x)}|geq epsilon$}
the above is a subset of the set $|f_n(x)-f(x)|geq epsilon/2$ obviously and thus is the
union of the set $|g_n(x)-g(x)|geq epsilon/2$ and hence the resukt holds.
Note here my measure is taken over an arbitrary set E. And this shows (1)
For (2) im drawing a blank and I don't have any idea how to even begin such a question. How do I 'handle' the zero function ?
measure-theory convergence
edited Nov 21 '18 at 10:13
saz
78.3k758123
asked Nov 21 '18 at 1:43
Jason Moore
607
add a comment |
(1)
If $f_n$ $longrightarrow$ $f$ in measure and $g_n$ $longrightarrow$ $g$ in measure then show that $f_n +g_n$ $longrightarrow$ $f+g$ in measure
(2)If $f_n longrightarrow theta$ in measure and $g_n longrightarrow theta$ in measure where $theta$ is the identically zero function on $Omega$ then show that $f_ng_n longrightarrow 0$
My attempt
For both cases I use the fact that $(Omega,Sigma,mu)$ is my measure space where everything is defined in the H.L Royden.
For (1) it follows from the relation $ {x in E |(f_n+g_n)(x)-(f+g)(x)}|geq epsilon$}
the above is a subset of the set $|f_n(x)-f(x)|geq epsilon/2$ obviously and thus is the
union of the set $|g_n(x)-g(x)|geq epsilon/2$ and hence the resukt holds.
Note here my measure is taken over an arbitrary set E. And this shows (1)
For (2) im drawing a blank and I don't have any idea how to even begin such a question. How do I 'handle' the zero function ?
measure-theory convergence
edited Nov 21 '18 at 10:13
saz
78.3k758123
asked Nov 21 '18 at 1:43
Jason Moore
607
I'm a bit puzzled about the title of your question. Since you already solved this part of the problem it would be more natural to to put the 2nd part in the title of your question...
– saz
Nov 21 '18 at 7:03
add a comment |
(1)
If $f_n$ $longrightarrow$ $f$ in measure and $g_n$ $longrightarrow$ $g$ in measure then show that $f_n +g_n$ $longrightarrow$ $f+g$ in measure
(2)If $f_n longrightarrow theta$ in measure and $g_n longrightarrow theta$ in measure where $theta$ is the identically zero function on $Omega$ then show that $f_ng_n longrightarrow 0$
My attempt
For both cases I use the fact that $(Omega,Sigma,mu)$ is my measure space where everything is defined in the H.L Royden.
For (1) it follows from the relation $ {x in E |(f_n+g_n)(x)-(f+g)(x)}|geq epsilon$}
the above is a subset of the set $|f_n(x)-f(x)|geq epsilon/2$ obviously and thus is the
union of the set $|g_n(x)-g(x)|geq epsilon/2$ and hence the resukt holds.
Note here my measure is taken over an arbitrary set E. And this shows (1)
For (2) im drawing a blank and I don't have any idea how to even begin such a question. How do I 'handle' the zero function ?
measure-theory convergence
edited Nov 21 '18 at 10:13
saz
78.3k758123
asked Nov 21 '18 at 1:43
Jason Moore
607
(1)
If $f_n$ $longrightarrow$ $f$ in measure and $g_n$ $longrightarrow$ $g$ in measure then show that $f_n +g_n$ $longrightarrow$ $f+g$ in measure
(2)If $f_n longrightarrow theta$ in measure and $g_n longrightarrow theta$ in measure where $theta$ is the identically zero function on $Omega$ then show that $f_ng_n longrightarrow 0$
My attempt
For both cases I use the fact that $(Omega,Sigma,mu)$ is my measure space where everything is defined in the H.L Royden.
For (1) it follows from the relation $ {x in E |(f_n+g_n)(x)-(f+g)(x)}|geq epsilon$}
the above is a subset of the set $|f_n(x)-f(x)|geq epsilon/2$ obviously and thus is the
union of the set $|g_n(x)-g(x)|geq epsilon/2$ and hence the resukt holds.
Note here my measure is taken over an arbitrary set E. And this shows (1)
For (2) im drawing a blank and I don't have any idea how to even begin such a question. How do I 'handle' the zero function ?
measure-theory convergence
measure-theory convergence
edited Nov 21 '18 at 10:13
saz
78.3k758123
asked Nov 21 '18 at 1:43
Jason Moore
607
edited Nov 21 '18 at 10:13
saz
78.3k758123
asked Nov 21 '18 at 1:43
Jason Moore
607
edited Nov 21 '18 at 10:13
saz
78.3k758123
edited Nov 21 '18 at 10:13
saz
78.3k758123
edited Nov 21 '18 at 10:13
saz
78.3k758123
78.3k758123
asked Nov 21 '18 at 1:43
Jason Moore
607
asked Nov 21 '18 at 1:43
Jason Moore
607
asked Nov 21 '18 at 1:43
Jason Moore
607
607
I'm a bit puzzled about the title of your question. Since you already solved this part of the problem it would be more natural to to put the 2nd part in the title of your question...
– saz
Nov 21 '18 at 7:03
add a comment |
I'm a bit puzzled about the title of your question. Since you already solved this part of the problem it would be more natural to to put the 2nd part in the title of your question...
– saz
Nov 21 '18 at 7:03
I'm a bit puzzled about the title of your question. Since you already solved this part of the problem it would be more natural to to put the 2nd part in the title of your question...
– saz
Nov 21 '18 at 7:03
I'm a bit puzzled about the title of your question. Since you already solved this part of the problem it would be more natural to to put the 2nd part in the title of your question...
– saz
Nov 21 '18 at 7:03
add a comment |
1 Answer
1
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Hints for (2):
- Let $(h_n)_{n in mathbb{N}}$ be a sequence of measurable functions. Prove that $h_n to 0$ in measure implies $h_n^2 to 0$ in measure.
- Use the first part of your problem and Step 1 to show that $(f_n+g_n)^2 to 0$ in measure and $(f_n-g_n)^2 to 0$ in measure.
- Use $$f_n cdot g_n = frac{1}{4} big( (f_n+g_n)^2- (f_n-g_n)^2 big)$$ and Step 1 to conclude that $f_n cdot g_n to 0$ in measure.
answered Nov 21 '18 at 7:01
saz
78.3k758123
Thanks very much this makes more sense to me now!
– Jason Moore
Nov 21 '18 at 14:27
@JasonMoore You are welcome.
– saz
Nov 21 '18 at 14:41
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
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oldest
votes
active
oldest
votes
Hints for (2):
- Let $(h_n)_{n in mathbb{N}}$ be a sequence of measurable functions. Prove that $h_n to 0$ in measure implies $h_n^2 to 0$ in measure.
- Use the first part of your problem and Step 1 to show that $(f_n+g_n)^2 to 0$ in measure and $(f_n-g_n)^2 to 0$ in measure.
- Use $$f_n cdot g_n = frac{1}{4} big( (f_n+g_n)^2- (f_n-g_n)^2 big)$$ and Step 1 to conclude that $f_n cdot g_n to 0$ in measure.
answered Nov 21 '18 at 7:01
saz
78.3k758123
Thanks very much this makes more sense to me now!
– Jason Moore
Nov 21 '18 at 14:27
@JasonMoore You are welcome.
– saz
Nov 21 '18 at 14:41
add a comment |
Hints for (2):
- Let $(h_n)_{n in mathbb{N}}$ be a sequence of measurable functions. Prove that $h_n to 0$ in measure implies $h_n^2 to 0$ in measure.
- Use the first part of your problem and Step 1 to show that $(f_n+g_n)^2 to 0$ in measure and $(f_n-g_n)^2 to 0$ in measure.
- Use $$f_n cdot g_n = frac{1}{4} big( (f_n+g_n)^2- (f_n-g_n)^2 big)$$ and Step 1 to conclude that $f_n cdot g_n to 0$ in measure.
answered Nov 21 '18 at 7:01
saz
78.3k758123
Thanks very much this makes more sense to me now!
– Jason Moore
Nov 21 '18 at 14:27
@JasonMoore You are welcome.
– saz
Nov 21 '18 at 14:41
add a comment |
Hints for (2):
- Let $(h_n)_{n in mathbb{N}}$ be a sequence of measurable functions. Prove that $h_n to 0$ in measure implies $h_n^2 to 0$ in measure.
- Use the first part of your problem and Step 1 to show that $(f_n+g_n)^2 to 0$ in measure and $(f_n-g_n)^2 to 0$ in measure.
- Use $$f_n cdot g_n = frac{1}{4} big( (f_n+g_n)^2- (f_n-g_n)^2 big)$$ and Step 1 to conclude that $f_n cdot g_n to 0$ in measure.
answered Nov 21 '18 at 7:01
saz
78.3k758123
Hints for (2):
- Let $(h_n)_{n in mathbb{N}}$ be a sequence of measurable functions. Prove that $h_n to 0$ in measure implies $h_n^2 to 0$ in measure.
- Use the first part of your problem and Step 1 to show that $(f_n+g_n)^2 to 0$ in measure and $(f_n-g_n)^2 to 0$ in measure.
- Use $$f_n cdot g_n = frac{1}{4} big( (f_n+g_n)^2- (f_n-g_n)^2 big)$$ and Step 1 to conclude that $f_n cdot g_n to 0$ in measure.
answered Nov 21 '18 at 7:01
saz
78.3k758123
answered Nov 21 '18 at 7:01
saz
78.3k758123
answered Nov 21 '18 at 7:01
saz
78.3k758123
answered Nov 21 '18 at 7:01
saz
78.3k758123
78.3k758123
Thanks very much this makes more sense to me now!
– Jason Moore
Nov 21 '18 at 14:27
@JasonMoore You are welcome.
– saz
Nov 21 '18 at 14:41
add a comment |
Thanks very much this makes more sense to me now!
– Jason Moore
Nov 21 '18 at 14:27
@JasonMoore You are welcome.
– saz
Nov 21 '18 at 14:41
Thanks very much this makes more sense to me now!
– Jason Moore
Nov 21 '18 at 14:27
Thanks very much this makes more sense to me now!
– Jason Moore
Nov 21 '18 at 14:27
@JasonMoore You are welcome.
– saz
Nov 21 '18 at 14:41
@JasonMoore You are welcome.
– saz
Nov 21 '18 at 14:41
add a comment |
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I'm a bit puzzled about the title of your question. Since you already solved this part of the problem it would be more natural to to put the 2nd part in the title of your question...
– saz
Nov 21 '18 at 7:03