Fdtxjr

The Two and the One

Is it unprofessional to ask if a job posting on GlassDoor is real?

Mathematical cryptic clues

What typically incentivizes a professor to change jobs to a lower ranking university?

Is it legal for company to use my work email to pretend I still work there?

Is it tax fraud for an individual to declare non-taxable revenue as taxable income? (US tax laws)

"to be prejudice towards/against someone" vs "to be prejudiced against/towards someone"

Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?

Smoothness of finite-dimensional functional calculus

Why Is Death Allowed In the Matrix?

How does one intimidate enemies without having the capacity for violence?

Prove that NP is closed under karp reduction?

Why can't I see bouncing of a switch on an oscilloscope?

What is the word for reserving something for yourself before others do?

What do you call a Matrix-like slowdown and camera movement effect?

A newer friend of my brother's gave him a load of baseball cards that are supposedly extremely valuable. Is this a scam?

How do I create uniquely male characters?

Why do I get two different answers for this counting problem?

Why doesn't H₄O²⁺ exist?

Approximately how much travel time was saved by the opening of the Suez Canal in 1869?

Theorems that impeded progress

Is it important to consider tone, melody, and musical form while writing a song?

LaTeX closing $ signs makes cursor jump

Which models of the Boeing 737 are still in production?

continuity of monotonically increasing function

f left continuous & strictly increasing; B Borel $implies$ f(B) Borel (or at least Lebesgue Measurable)?Discontinuities of a monotonic function (Baby Rudin)If two functions are both increasing and equal almost everywhere, do they have the same discontinuous points?Doubt in discontinuities of a monotonic function.Bivariate function monotone in each variable $Rightarrow$ continuous a.e.?Characteristics of a monotonic function on a closed interval$ f : D rightarrow mathbbR $ monotone increasing. Cardinality of discontinuity points?Inverse of function, monotone on interval is continuousthe continuity of monotone additive function on $mathbbR$Monotone Functions and Continuities

1

$begingroup$

Let $f:[0,1]to[0,1]$ be monotonically increasing. Which of the following statements is/are true?

1. $f$ must be continuous at all but finitely many points in $[0,1]$


2. $f$ must be continuous at all but countably many points in $[0,1]$


3. $f$ must be Riemann integrable


4. $f$ must be Lebesgue integrable

I know that set of discontinuities of monotonic function is at most countable so 1 is not true.

If the set of discontinuities is at most countable then how to conclude for the second option that is set of continuities of monotone function. Here domain is [0,1] .So the set of continuities can be uncountable?

Monotone function on $[a,b]$ is Riemann integrable and hence Lebesgue integrable hence option 3 and 4 are correct.

monotone-functions

share|cite|improve this question

edited May 12 '17 at 9:08

dipali mali

asked May 12 '17 at 8:53

dipali malidipali mali

1715

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A monotonic function can only have discontinuity of the first kind
  $endgroup$
  – Li Chun Min
  May 12 '17 at 8:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  , which is at most countable
  $endgroup$
  – Li Chun Min
  May 12 '17 at 8:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f$ is Riemann integrable iff $f$ is continuous almost everywhere on $[0,1]$. Therefore since a monotone function is Riemann integrable 2 must be correct.
  $endgroup$
  – Janitha357
  May 12 '17 at 9:00
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $f:[0,1]to[0,1]$ be monotonically increasing. Which of the following statements is/are true?

1. $f$ must be continuous at all but finitely many points in $[0,1]$


2. $f$ must be continuous at all but countably many points in $[0,1]$


3. $f$ must be Riemann integrable


4. $f$ must be Lebesgue integrable

I know that set of discontinuities of monotonic function is at most countable so 1 is not true.

If the set of discontinuities is at most countable then how to conclude for the second option that is set of continuities of monotone function. Here domain is [0,1] .So the set of continuities can be uncountable?

Monotone function on $[a,b]$ is Riemann integrable and hence Lebesgue integrable hence option 3 and 4 are correct.

monotone-functions

share|cite|improve this question

edited May 12 '17 at 9:08

dipali mali

asked May 12 '17 at 8:53

dipali malidipali mali

1715

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A monotonic function can only have discontinuity of the first kind
  $endgroup$
  – Li Chun Min
  May 12 '17 at 8:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  , which is at most countable
  $endgroup$
  – Li Chun Min
  May 12 '17 at 8:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f$ is Riemann integrable iff $f$ is continuous almost everywhere on $[0,1]$. Therefore since a monotone function is Riemann integrable 2 must be correct.
  $endgroup$
  – Janitha357
  May 12 '17 at 9:00
  

  
  
  
  

add a comment | 

$begingroup$

Let $f:[0,1]to[0,1]$ be monotonically increasing. Which of the following statements is/are true?

1. $f$ must be continuous at all but finitely many points in $[0,1]$


2. $f$ must be continuous at all but countably many points in $[0,1]$


3. $f$ must be Riemann integrable


4. $f$ must be Lebesgue integrable

I know that set of discontinuities of monotonic function is at most countable so 1 is not true.

If the set of discontinuities is at most countable then how to conclude for the second option that is set of continuities of monotone function. Here domain is [0,1] .So the set of continuities can be uncountable?

Monotone function on $[a,b]$ is Riemann integrable and hence Lebesgue integrable hence option 3 and 4 are correct.

monotone-functions

share|cite|improve this question

edited May 12 '17 at 9:08

dipali mali

asked May 12 '17 at 8:53

dipali malidipali mali

1715

$endgroup$

share|cite|improve this question

edited May 12 '17 at 9:08

dipali mali

asked May 12 '17 at 8:53

dipali malidipali mali

1715

share|cite|improve this question

edited May 12 '17 at 9:08

dipali mali

asked May 12 '17 at 8:53

dipali malidipali mali

1715

asked May 12 '17 at 8:53

dipali malidipali mali

1715

asked May 12 '17 at 8:53

dipali malidipali mali

1715

1 Answer
1

active

oldest

votes

0

$begingroup$

Let $f(x+)=lim_yrightarrow x^+f(y)$ and $f(x-)=lim_yrightarrow x^-f(y)$. Note that $f(x+)geq f(x-)~forall x$ as $f$ is increasing. Then let $D_n=x:f(x+)-f(x-)geqfrac1n$. As $f:[0,1]rightarrow[0,1]$, hence $|D_n|leq n$. Also the set of all discontinuities, $D=cup_n=1^inftyD_n$, which is a countable union of finite sets, and hence is countable.

Also Lebesgue measure of a countable set is $0$. Hence Lebesgue measurable.

share|cite|improve this answer

answered May 12 '17 at 9:06

Abishanka SahaAbishanka Saha

7,90511022

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  second option is correct?
  $endgroup$
  – dipali mali
  May 13 '17 at 11:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes it is correct
  $endgroup$
  – Abishanka Saha
  May 13 '17 at 11:57
  

  
  
  
  

add a comment | 

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',
autoActivateHeartbeat: false,
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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2277653%2fcontinuity-of-monotonically-increasing-function%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

Let $f(x+)=lim_yrightarrow x^+f(y)$ and $f(x-)=lim_yrightarrow x^-f(y)$. Note that $f(x+)geq f(x-)~forall x$ as $f$ is increasing. Then let $D_n=x:f(x+)-f(x-)geqfrac1n$. As $f:[0,1]rightarrow[0,1]$, hence $|D_n|leq n$. Also the set of all discontinuities, $D=cup_n=1^inftyD_n$, which is a countable union of finite sets, and hence is countable.

Also Lebesgue measure of a countable set is $0$. Hence Lebesgue measurable.

share|cite|improve this answer

answered May 12 '17 at 9:06

Abishanka SahaAbishanka Saha

7,90511022

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  second option is correct?
  $endgroup$
  – dipali mali
  May 13 '17 at 11:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes it is correct
  $endgroup$
  – Abishanka Saha
  May 13 '17 at 11:57
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $f(x+)=lim_yrightarrow x^+f(y)$ and $f(x-)=lim_yrightarrow x^-f(y)$. Note that $f(x+)geq f(x-)~forall x$ as $f$ is increasing. Then let $D_n=x:f(x+)-f(x-)geqfrac1n$. As $f:[0,1]rightarrow[0,1]$, hence $|D_n|leq n$. Also the set of all discontinuities, $D=cup_n=1^inftyD_n$, which is a countable union of finite sets, and hence is countable.

Also Lebesgue measure of a countable set is $0$. Hence Lebesgue measurable.

share|cite|improve this answer

answered May 12 '17 at 9:06

Abishanka SahaAbishanka Saha

7,90511022

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  second option is correct?
  $endgroup$
  – dipali mali
  May 13 '17 at 11:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes it is correct
  $endgroup$
  – Abishanka Saha
  May 13 '17 at 11:57
  

  
  
  
  

add a comment | 

$begingroup$

Let $f(x+)=lim_yrightarrow x^+f(y)$ and $f(x-)=lim_yrightarrow x^-f(y)$. Note that $f(x+)geq f(x-)~forall x$ as $f$ is increasing. Then let $D_n=x:f(x+)-f(x-)geqfrac1n$. As $f:[0,1]rightarrow[0,1]$, hence $|D_n|leq n$. Also the set of all discontinuities, $D=cup_n=1^inftyD_n$, which is a countable union of finite sets, and hence is countable.

Also Lebesgue measure of a countable set is $0$. Hence Lebesgue measurable.

share|cite|improve this answer

answered May 12 '17 at 9:06

Abishanka SahaAbishanka Saha

7,90511022

$endgroup$

share|cite|improve this answer

answered May 12 '17 at 9:06

Abishanka SahaAbishanka Saha

7,90511022

answered May 12 '17 at 9:06

Abishanka SahaAbishanka Saha

7,90511022

answered May 12 '17 at 9:06

Abishanka SahaAbishanka Saha

7,90511022