Manipulations with convergence a.e.
$begingroup$
Let functions $f_n$ be measurable, $n in N$, $f_nrightarrow f$ almost everywhere. Prove that $operatorname{arctg}f_n rightarrow operatorname{arctg}f$ almost everywhere.
Honestly speaking, we have had so many theorems on different types of convergence and relationships between them that I find it very hard to mix them in order to solve problems.
First of all, we are given that the measure of the set of points for which $f_n$ does not converge to $f$ is zero.
How should I think about the problem next? How do we show that $arctg$ preserves convergence?
I have my lecture notes, but when it comes to practice I often feel helpless and do not know how to approach the problem. Probably, you could recommend some good problem book with solved examples.
Thank you
continuity almost-everywhere pointwise-convergence
edited Dec 25 '18 at 12:59
Hayk
2,6271214
asked Dec 25 '18 at 12:50
Don DraperDon Draper
87110
$endgroup$
add a comment |
$begingroup$
Let functions $f_n$ be measurable, $n in N$, $f_nrightarrow f$ almost everywhere. Prove that $operatorname{arctg}f_n rightarrow operatorname{arctg}f$ almost everywhere.
Honestly speaking, we have had so many theorems on different types of convergence and relationships between them that I find it very hard to mix them in order to solve problems.
First of all, we are given that the measure of the set of points for which $f_n$ does not converge to $f$ is zero.
How should I think about the problem next? How do we show that $arctg$ preserves convergence?
I have my lecture notes, but when it comes to practice I often feel helpless and do not know how to approach the problem. Probably, you could recommend some good problem book with solved examples.
Thank you
continuity almost-everywhere pointwise-convergence
edited Dec 25 '18 at 12:59
Hayk
2,6271214
asked Dec 25 '18 at 12:50
Don DraperDon Draper
87110
$endgroup$
1
$begingroup$
use the continuity of $arctg$, namely fix a point $x$ where $f_n(x) to f(x)$ and prove that $arctg f_n(x) to arctg f(x)$ for that $x$ using the continuity of $arctg$ at the point $f(x)$.
$endgroup$
– Hayk
Dec 25 '18 at 12:56
add a comment |
$begingroup$
Let functions $f_n$ be measurable, $n in N$, $f_nrightarrow f$ almost everywhere. Prove that $operatorname{arctg}f_n rightarrow operatorname{arctg}f$ almost everywhere.
Honestly speaking, we have had so many theorems on different types of convergence and relationships between them that I find it very hard to mix them in order to solve problems.
First of all, we are given that the measure of the set of points for which $f_n$ does not converge to $f$ is zero.
How should I think about the problem next? How do we show that $arctg$ preserves convergence?
I have my lecture notes, but when it comes to practice I often feel helpless and do not know how to approach the problem. Probably, you could recommend some good problem book with solved examples.
Thank you
continuity almost-everywhere pointwise-convergence
edited Dec 25 '18 at 12:59
Hayk
2,6271214
asked Dec 25 '18 at 12:50
Don DraperDon Draper
87110
$endgroup$
Let functions $f_n$ be measurable, $n in N$, $f_nrightarrow f$ almost everywhere. Prove that $operatorname{arctg}f_n rightarrow operatorname{arctg}f$ almost everywhere.
Honestly speaking, we have had so many theorems on different types of convergence and relationships between them that I find it very hard to mix them in order to solve problems.
First of all, we are given that the measure of the set of points for which $f_n$ does not converge to $f$ is zero.
How should I think about the problem next? How do we show that $arctg$ preserves convergence?
I have my lecture notes, but when it comes to practice I often feel helpless and do not know how to approach the problem. Probably, you could recommend some good problem book with solved examples.
Thank you
continuity almost-everywhere pointwise-convergence
continuity almost-everywhere pointwise-convergence
edited Dec 25 '18 at 12:59
Hayk
2,6271214
asked Dec 25 '18 at 12:50
Don DraperDon Draper
87110
edited Dec 25 '18 at 12:59
Hayk
2,6271214
asked Dec 25 '18 at 12:50
Don DraperDon Draper
87110
edited Dec 25 '18 at 12:59
Hayk
2,6271214
edited Dec 25 '18 at 12:59
Hayk
2,6271214
edited Dec 25 '18 at 12:59
Hayk
2,6271214
2,6271214
asked Dec 25 '18 at 12:50
Don DraperDon Draper
87110
asked Dec 25 '18 at 12:50
Don DraperDon Draper
87110
asked Dec 25 '18 at 12:50
Don DraperDon Draper
87110
87110
1
$begingroup$
use the continuity of $arctg$, namely fix a point $x$ where $f_n(x) to f(x)$ and prove that $arctg f_n(x) to arctg f(x)$ for that $x$ using the continuity of $arctg$ at the point $f(x)$.
$endgroup$
– Hayk
Dec 25 '18 at 12:56
add a comment |
1
$begingroup$
use the continuity of $arctg$, namely fix a point $x$ where $f_n(x) to f(x)$ and prove that $arctg f_n(x) to arctg f(x)$ for that $x$ using the continuity of $arctg$ at the point $f(x)$.
$endgroup$
– Hayk
Dec 25 '18 at 12:56
1
1
$begingroup$
use the continuity of $arctg$, namely fix a point $x$ where $f_n(x) to f(x)$ and prove that $arctg f_n(x) to arctg f(x)$ for that $x$ using the continuity of $arctg$ at the point $f(x)$.
$endgroup$
– Hayk
Dec 25 '18 at 12:56
$begingroup$
use the continuity of $arctg$, namely fix a point $x$ where $f_n(x) to f(x)$ and prove that $arctg f_n(x) to arctg f(x)$ for that $x$ using the continuity of $arctg$ at the point $f(x)$.
$endgroup$
– Hayk
Dec 25 '18 at 12:56
add a comment |
1 Answer
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$begingroup$
Suppose that $g : E subset mathbb{R}^l to mathbb{R}$ is continuous, and let $(f_n)_n :mathbb{R}^k to E$ converge almost everywhere to some function $f : mathbb{R}^k to E$. Now, by continuity, if $x in mathbb{R}^k$ is such that $f_n(x) to f(x)$, then $gf_n(x) to gf(x)$. By the contrapositive, then, if $gf_n(x) not to gf(x)$ we have that $f_n(x) not to f(x)$. That is,
$$
{x : gf_n(x) to gf(x)}^c subset {x : f_n(x) to f(x)}^c
$$
and so
$$
mu({x : gf_n(x) to gf(x)}^c) leq mu({x : f_n(x) to f(x)}^c) = 0.
$$
Hence $gf_n to gf$ a.e.
answered Dec 25 '18 at 13:00
Guido A.Guido A.
7,8701730
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add a comment |
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$begingroup$
Suppose that $g : E subset mathbb{R}^l to mathbb{R}$ is continuous, and let $(f_n)_n :mathbb{R}^k to E$ converge almost everywhere to some function $f : mathbb{R}^k to E$. Now, by continuity, if $x in mathbb{R}^k$ is such that $f_n(x) to f(x)$, then $gf_n(x) to gf(x)$. By the contrapositive, then, if $gf_n(x) not to gf(x)$ we have that $f_n(x) not to f(x)$. That is,
$$
{x : gf_n(x) to gf(x)}^c subset {x : f_n(x) to f(x)}^c
$$
and so
$$
mu({x : gf_n(x) to gf(x)}^c) leq mu({x : f_n(x) to f(x)}^c) = 0.
$$
Hence $gf_n to gf$ a.e.
answered Dec 25 '18 at 13:00
Guido A.Guido A.
7,8701730
$endgroup$
add a comment |
$begingroup$
Suppose that $g : E subset mathbb{R}^l to mathbb{R}$ is continuous, and let $(f_n)_n :mathbb{R}^k to E$ converge almost everywhere to some function $f : mathbb{R}^k to E$. Now, by continuity, if $x in mathbb{R}^k$ is such that $f_n(x) to f(x)$, then $gf_n(x) to gf(x)$. By the contrapositive, then, if $gf_n(x) not to gf(x)$ we have that $f_n(x) not to f(x)$. That is,
$$
{x : gf_n(x) to gf(x)}^c subset {x : f_n(x) to f(x)}^c
$$
and so
$$
mu({x : gf_n(x) to gf(x)}^c) leq mu({x : f_n(x) to f(x)}^c) = 0.
$$
Hence $gf_n to gf$ a.e.
answered Dec 25 '18 at 13:00
Guido A.Guido A.
7,8701730
$endgroup$
add a comment |
$begingroup$
Suppose that $g : E subset mathbb{R}^l to mathbb{R}$ is continuous, and let $(f_n)_n :mathbb{R}^k to E$ converge almost everywhere to some function $f : mathbb{R}^k to E$. Now, by continuity, if $x in mathbb{R}^k$ is such that $f_n(x) to f(x)$, then $gf_n(x) to gf(x)$. By the contrapositive, then, if $gf_n(x) not to gf(x)$ we have that $f_n(x) not to f(x)$. That is,
$$
{x : gf_n(x) to gf(x)}^c subset {x : f_n(x) to f(x)}^c
$$
and so
$$
mu({x : gf_n(x) to gf(x)}^c) leq mu({x : f_n(x) to f(x)}^c) = 0.
$$
Hence $gf_n to gf$ a.e.
answered Dec 25 '18 at 13:00
Guido A.Guido A.
7,8701730
$endgroup$
Suppose that $g : E subset mathbb{R}^l to mathbb{R}$ is continuous, and let $(f_n)_n :mathbb{R}^k to E$ converge almost everywhere to some function $f : mathbb{R}^k to E$. Now, by continuity, if $x in mathbb{R}^k$ is such that $f_n(x) to f(x)$, then $gf_n(x) to gf(x)$. By the contrapositive, then, if $gf_n(x) not to gf(x)$ we have that $f_n(x) not to f(x)$. That is,
$$
{x : gf_n(x) to gf(x)}^c subset {x : f_n(x) to f(x)}^c
$$
and so
$$
mu({x : gf_n(x) to gf(x)}^c) leq mu({x : f_n(x) to f(x)}^c) = 0.
$$
Hence $gf_n to gf$ a.e.
answered Dec 25 '18 at 13:00
Guido A.Guido A.
7,8701730
edited Dec 25 '18 at 13:06
answered Dec 25 '18 at 13:00
Guido A.Guido A.
7,8701730
answered Dec 25 '18 at 13:00
Guido A.Guido A.
7,8701730
answered Dec 25 '18 at 13:00
Guido A.Guido A.
7,8701730
7,8701730
add a comment |
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$begingroup$
use the continuity of $arctg$, namely fix a point $x$ where $f_n(x) to f(x)$ and prove that $arctg f_n(x) to arctg f(x)$ for that $x$ using the continuity of $arctg$ at the point $f(x)$.
$endgroup$
– Hayk
Dec 25 '18 at 12:56