If $f_n longrightarrow f$ almost everywhere and $g_n longrightarrow g$ almost everywhere
up vote
0
down vote
favorite
1 The Riemann function $f:[-1,1]longrightarrow mathbb{R}$ defined as
$$
f(x)=
begin{cases}
0, &textrm{if }x notin mathbb{Q}cap[-1,1]textrm{ or }x =0\
frac{1}{q} , &textrm{if }xin mathbb{Q} cap[-1,1]textrm{ and }x =frac{p}{q}neq 0
end{cases}
$$
Where $x =frac{p}{q}$ is an irreducible fraction and q is non zero ($qneq o)$
Does the Lebesgue Criterion for Riemann Integrability hold for this function?
[2] If $f_n longrightarrow f$ almost everywhere and $g_n longrightarrow g$ almost everywhere where $f,g$ are both finite valued, then show that
$f_n+g_n longrightarrow f+g$ almost everywhere And
$f_ng_n longrightarrow fg$ almost everywhere
MY ATTEMPT
1 YES!
$f$ is discontinuous at any rational point $x =frac{p}{q}$ $(q neq0)$ and continuous at any irrational point in [-1,1] and f is bounded.
Since $A= mathbb{Q} cap[-1,1]$ -{0}the set of discontinuous points of f. Hence $m(A)=0$
Therefore by Lebesgue Criterion for Riemann integrability $f$ is Riemann integrable on [-1,1] Hence $f$ is Lebesgue Integrable and
$int^{1}_{-1} f(x) dx = int_{[-1,1]} fdm=0$ because $f=0$ almost everywhere
IS this correct?
[2] I'm not sure how to approach this question. I've tried using THIS
but I am unable to see the connection if there is any at all. Never did a question like this so i'm unsure as to how to begin.
measure-theory lebesgue-integral lebesgue-measure almost-everywhere measurable-functions
edited Nov 21 at 18:37
MPW
29.8k11956
asked Nov 21 at 18:31
Jason Moore
607
add a comment |
up vote
0
down vote
favorite
1 The Riemann function $f:[-1,1]longrightarrow mathbb{R}$ defined as
$$
f(x)=
begin{cases}
0, &textrm{if }x notin mathbb{Q}cap[-1,1]textrm{ or }x =0\
frac{1}{q} , &textrm{if }xin mathbb{Q} cap[-1,1]textrm{ and }x =frac{p}{q}neq 0
end{cases}
$$
Where $x =frac{p}{q}$ is an irreducible fraction and q is non zero ($qneq o)$
Does the Lebesgue Criterion for Riemann Integrability hold for this function?
[2] If $f_n longrightarrow f$ almost everywhere and $g_n longrightarrow g$ almost everywhere where $f,g$ are both finite valued, then show that
$f_n+g_n longrightarrow f+g$ almost everywhere And
$f_ng_n longrightarrow fg$ almost everywhere
MY ATTEMPT
1 YES!
$f$ is discontinuous at any rational point $x =frac{p}{q}$ $(q neq0)$ and continuous at any irrational point in [-1,1] and f is bounded.
Since $A= mathbb{Q} cap[-1,1]$ -{0}the set of discontinuous points of f. Hence $m(A)=0$
Therefore by Lebesgue Criterion for Riemann integrability $f$ is Riemann integrable on [-1,1] Hence $f$ is Lebesgue Integrable and
$int^{1}_{-1} f(x) dx = int_{[-1,1]} fdm=0$ because $f=0$ almost everywhere
IS this correct?
[2] I'm not sure how to approach this question. I've tried using THIS
but I am unable to see the connection if there is any at all. Never did a question like this so i'm unsure as to how to begin.
measure-theory lebesgue-integral lebesgue-measure almost-everywhere measurable-functions
edited Nov 21 at 18:37
MPW
29.8k11956
asked Nov 21 at 18:31
Jason Moore
607
Don't link documents. For (2) just take a null set $N$ such that outside of it both $f_n$ and $g_n$ converge. You are reduced to sequences of real (or complex) numbers.
– Will M.
Nov 21 at 18:34
You should require $q>0$ or the function is not well-defined (because $frac pq=frac{-p}{-q}$). I already added the requirement that $pneq 0$ for you.
– MPW
Nov 21 at 18:41
Ok i've fixed part (1) but i'm still pretty confused about part (2). Anything else you can tell me mate?
– Jason Moore
Nov 22 at 13:53
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
1 The Riemann function $f:[-1,1]longrightarrow mathbb{R}$ defined as
$$
f(x)=
begin{cases}
0, &textrm{if }x notin mathbb{Q}cap[-1,1]textrm{ or }x =0\
frac{1}{q} , &textrm{if }xin mathbb{Q} cap[-1,1]textrm{ and }x =frac{p}{q}neq 0
end{cases}
$$
Where $x =frac{p}{q}$ is an irreducible fraction and q is non zero ($qneq o)$
Does the Lebesgue Criterion for Riemann Integrability hold for this function?
[2] If $f_n longrightarrow f$ almost everywhere and $g_n longrightarrow g$ almost everywhere where $f,g$ are both finite valued, then show that
$f_n+g_n longrightarrow f+g$ almost everywhere And
$f_ng_n longrightarrow fg$ almost everywhere
MY ATTEMPT
1 YES!
$f$ is discontinuous at any rational point $x =frac{p}{q}$ $(q neq0)$ and continuous at any irrational point in [-1,1] and f is bounded.
Since $A= mathbb{Q} cap[-1,1]$ -{0}the set of discontinuous points of f. Hence $m(A)=0$
Therefore by Lebesgue Criterion for Riemann integrability $f$ is Riemann integrable on [-1,1] Hence $f$ is Lebesgue Integrable and
$int^{1}_{-1} f(x) dx = int_{[-1,1]} fdm=0$ because $f=0$ almost everywhere
IS this correct?
[2] I'm not sure how to approach this question. I've tried using THIS
but I am unable to see the connection if there is any at all. Never did a question like this so i'm unsure as to how to begin.
measure-theory lebesgue-integral lebesgue-measure almost-everywhere measurable-functions
edited Nov 21 at 18:37
MPW
29.8k11956
asked Nov 21 at 18:31
Jason Moore
607
1 The Riemann function $f:[-1,1]longrightarrow mathbb{R}$ defined as
$$
f(x)=
begin{cases}
0, &textrm{if }x notin mathbb{Q}cap[-1,1]textrm{ or }x =0\
frac{1}{q} , &textrm{if }xin mathbb{Q} cap[-1,1]textrm{ and }x =frac{p}{q}neq 0
end{cases}
$$
Where $x =frac{p}{q}$ is an irreducible fraction and q is non zero ($qneq o)$
Does the Lebesgue Criterion for Riemann Integrability hold for this function?
[2] If $f_n longrightarrow f$ almost everywhere and $g_n longrightarrow g$ almost everywhere where $f,g$ are both finite valued, then show that
$f_n+g_n longrightarrow f+g$ almost everywhere And
$f_ng_n longrightarrow fg$ almost everywhere
MY ATTEMPT
1 YES!
$f$ is discontinuous at any rational point $x =frac{p}{q}$ $(q neq0)$ and continuous at any irrational point in [-1,1] and f is bounded.
Since $A= mathbb{Q} cap[-1,1]$ -{0}the set of discontinuous points of f. Hence $m(A)=0$
Therefore by Lebesgue Criterion for Riemann integrability $f$ is Riemann integrable on [-1,1] Hence $f$ is Lebesgue Integrable and
$int^{1}_{-1} f(x) dx = int_{[-1,1]} fdm=0$ because $f=0$ almost everywhere
IS this correct?
[2] I'm not sure how to approach this question. I've tried using THIS
but I am unable to see the connection if there is any at all. Never did a question like this so i'm unsure as to how to begin.
measure-theory lebesgue-integral lebesgue-measure almost-everywhere measurable-functions
measure-theory lebesgue-integral lebesgue-measure almost-everywhere measurable-functions
edited Nov 21 at 18:37
MPW
29.8k11956
asked Nov 21 at 18:31
Jason Moore
607
edited Nov 21 at 18:37
MPW
29.8k11956
asked Nov 21 at 18:31
Jason Moore
607
edited Nov 21 at 18:37
MPW
29.8k11956
edited Nov 21 at 18:37
MPW
29.8k11956
edited Nov 21 at 18:37
MPW
29.8k11956
29.8k11956
asked Nov 21 at 18:31
Jason Moore
607
asked Nov 21 at 18:31
Jason Moore
607
asked Nov 21 at 18:31
Jason Moore
607
607
Don't link documents. For (2) just take a null set $N$ such that outside of it both $f_n$ and $g_n$ converge. You are reduced to sequences of real (or complex) numbers.
– Will M.
Nov 21 at 18:34
You should require $q>0$ or the function is not well-defined (because $frac pq=frac{-p}{-q}$). I already added the requirement that $pneq 0$ for you.
– MPW
Nov 21 at 18:41
Ok i've fixed part (1) but i'm still pretty confused about part (2). Anything else you can tell me mate?
– Jason Moore
Nov 22 at 13:53
add a comment |
Don't link documents. For (2) just take a null set $N$ such that outside of it both $f_n$ and $g_n$ converge. You are reduced to sequences of real (or complex) numbers.
– Will M.
Nov 21 at 18:34
You should require $q>0$ or the function is not well-defined (because $frac pq=frac{-p}{-q}$). I already added the requirement that $pneq 0$ for you.
– MPW
Nov 21 at 18:41
Ok i've fixed part (1) but i'm still pretty confused about part (2). Anything else you can tell me mate?
– Jason Moore
Nov 22 at 13:53
Don't link documents. For (2) just take a null set $N$ such that outside of it both $f_n$ and $g_n$ converge. You are reduced to sequences of real (or complex) numbers.
– Will M.
Nov 21 at 18:34
Don't link documents. For (2) just take a null set $N$ such that outside of it both $f_n$ and $g_n$ converge. You are reduced to sequences of real (or complex) numbers.
– Will M.
Nov 21 at 18:34
You should require $q>0$ or the function is not well-defined (because $frac pq=frac{-p}{-q}$). I already added the requirement that $pneq 0$ for you.
– MPW
Nov 21 at 18:41
You should require $q>0$ or the function is not well-defined (because $frac pq=frac{-p}{-q}$). I already added the requirement that $pneq 0$ for you.
– MPW
Nov 21 at 18:41
Ok i've fixed part (1) but i'm still pretty confused about part (2). Anything else you can tell me mate?
– Jason Moore
Nov 22 at 13:53
Ok i've fixed part (1) but i'm still pretty confused about part (2). Anything else you can tell me mate?
– Jason Moore
Nov 22 at 13:53
add a comment |
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008145%2fif-f-n-longrightarrow-f-almost-everywhere-and-g-n-longrightarrow-g-almost%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008145%2fif-f-n-longrightarrow-f-almost-everywhere-and-g-n-longrightarrow-g-almost%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Don't link documents. For (2) just take a null set $N$ such that outside of it both $f_n$ and $g_n$ converge. You are reduced to sequences of real (or complex) numbers.
– Will M.
Nov 21 at 18:34
You should require $q>0$ or the function is not well-defined (because $frac pq=frac{-p}{-q}$). I already added the requirement that $pneq 0$ for you.
– MPW
Nov 21 at 18:41
Ok i've fixed part (1) but i'm still pretty confused about part (2). Anything else you can tell me mate?
– Jason Moore
Nov 22 at 13:53