Composite of non Riemann integrable functions can be Riemann integrable?












1












$begingroup$


(1) Let $f,g$ be not Riemann integrable on $[a,b]$, and the range of $f$ is $[a,b]$ also. Can we find an example such that $gcirc f(x)=g(f(x))$ is Riemann integrable on $[a,b]$?



(2) Let $f,g$ be Riemann integrable on $[a,b]$, and the range of $f$ is $[a,b]$ also. Can we show that $gcirc f$ is Riemann integrable on　 $[a,b]$ also?






calculus integration







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  • $begingroup$
    Also, please avoid multiple questions in a post.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 5:08


















1












$begingroup$


(1) Let $f,g$ be not Riemann integrable on $[a,b]$, and the range of $f$ is $[a,b]$ also. Can we find an example such that $gcirc f(x)=g(f(x))$ is Riemann integrable on $[a,b]$?



(2) Let $f,g$ be Riemann integrable on $[a,b]$, and the range of $f$ is $[a,b]$ also. Can we show that $gcirc f$ is Riemann integrable on　 $[a,b]$ also?






calculus integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    Also, please avoid multiple questions in a post.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 5:08
















1












1








1





$begingroup$


(1) Let $f,g$ be not Riemann integrable on $[a,b]$, and the range of $f$ is $[a,b]$ also. Can we find an example such that $gcirc f(x)=g(f(x))$ is Riemann integrable on $[a,b]$?



(2) Let $f,g$ be Riemann integrable on $[a,b]$, and the range of $f$ is $[a,b]$ also. Can we show that $gcirc f$ is Riemann integrable on　 $[a,b]$ also?






calculus integration







share|cite|improve this question









$endgroup$




(1) Let $f,g$ be not Riemann integrable on $[a,b]$, and the range of $f$ is $[a,b]$ also. Can we find an example such that $gcirc f(x)=g(f(x))$ is Riemann integrable on $[a,b]$?



(2) Let $f,g$ be Riemann integrable on $[a,b]$, and the range of $f$ is $[a,b]$ also. Can we show that $gcirc f$ is Riemann integrable on　 $[a,b]$ also?






calculus integration




calculus integration






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share|cite|improve this question










asked Dec 8 '18 at 5:00









xlddxldd

1,330510




1,330510












  • $begingroup$
    Also, please avoid multiple questions in a post.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 5:08




















  • $begingroup$
    Also, please avoid multiple questions in a post.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 5:08


















$begingroup$
Also, please avoid multiple questions in a post.
$endgroup$
– Kemono Chen
Dec 8 '18 at 5:08






$begingroup$
Also, please avoid multiple questions in a post.
$endgroup$
– Kemono Chen
Dec 8 '18 at 5:08












1 Answer
1






active

oldest

votes


















2












$begingroup$

For question 1, $$f(x)=g(x)=
left{
begin{array}{ll}
x, &xin mathbb{Q} \
1-x,&xnotin mathbb{Q} \
end{array}
right.
\$$and $[a,b]=[0,1]$.



For question 2, see here.
In order to change the range of $f$, alter it to
$$ f(x) = begin{cases} 1/q & text{ for }x=p/qtext{ and $0le xle1/2$} \ 0 & text{ for } x notin mathbb{Q}text{ and $0le xle1/2$}\
2x-1 & text{others} end{cases}
$$ and the $g$ desired is the $f$ in my link.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    question 2, there is another condition: the range of $f$ is $[a,b]$.
    $endgroup$
    – xldd
    Dec 8 '18 at 7:29










  • $begingroup$
    @xldd edited. Please check.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 7:53










  • $begingroup$
    @Chen What is the outer function and the inner function? It really troubles me.
    $endgroup$
    – xldd
    Dec 8 '18 at 10:28












  • $begingroup$
    In math.stackexchange.com/questions/1060834/…, the inner function is the Riemann function, whose range is not all of $[0,1]$, and the $f$ is the outer function, which seems you have changed to be here.
    $endgroup$
    – xldd
    Dec 8 '18 at 10:30






  • 1




    $begingroup$
    It is not hard to alter the inner function. Try to do it yourself. Anyway I edited it for correctness.
    $endgroup$
    – Kemono Chen
    Dec 10 '18 at 2:45











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

For question 1, $$f(x)=g(x)=
left{
begin{array}{ll}
x, &xin mathbb{Q} \
1-x,&xnotin mathbb{Q} \
end{array}
right.
\$$and $[a,b]=[0,1]$.



For question 2, see here.
In order to change the range of $f$, alter it to
$$ f(x) = begin{cases} 1/q & text{ for }x=p/qtext{ and $0le xle1/2$} \ 0 & text{ for } x notin mathbb{Q}text{ and $0le xle1/2$}\
2x-1 & text{others} end{cases}
$$ and the $g$ desired is the $f$ in my link.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    question 2, there is another condition: the range of $f$ is $[a,b]$.
    $endgroup$
    – xldd
    Dec 8 '18 at 7:29










  • $begingroup$
    @xldd edited. Please check.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 7:53










  • $begingroup$
    @Chen What is the outer function and the inner function? It really troubles me.
    $endgroup$
    – xldd
    Dec 8 '18 at 10:28












  • $begingroup$
    In math.stackexchange.com/questions/1060834/…, the inner function is the Riemann function, whose range is not all of $[0,1]$, and the $f$ is the outer function, which seems you have changed to be here.
    $endgroup$
    – xldd
    Dec 8 '18 at 10:30






  • 1




    $begingroup$
    It is not hard to alter the inner function. Try to do it yourself. Anyway I edited it for correctness.
    $endgroup$
    – Kemono Chen
    Dec 10 '18 at 2:45
















2












$begingroup$

For question 1, $$f(x)=g(x)=
left{
begin{array}{ll}
x, &xin mathbb{Q} \
1-x,&xnotin mathbb{Q} \
end{array}
right.
\$$and $[a,b]=[0,1]$.



For question 2, see here.
In order to change the range of $f$, alter it to
$$ f(x) = begin{cases} 1/q & text{ for }x=p/qtext{ and $0le xle1/2$} \ 0 & text{ for } x notin mathbb{Q}text{ and $0le xle1/2$}\
2x-1 & text{others} end{cases}
$$ and the $g$ desired is the $f$ in my link.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    question 2, there is another condition: the range of $f$ is $[a,b]$.
    $endgroup$
    – xldd
    Dec 8 '18 at 7:29










  • $begingroup$
    @xldd edited. Please check.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 7:53










  • $begingroup$
    @Chen What is the outer function and the inner function? It really troubles me.
    $endgroup$
    – xldd
    Dec 8 '18 at 10:28












  • $begingroup$
    In math.stackexchange.com/questions/1060834/…, the inner function is the Riemann function, whose range is not all of $[0,1]$, and the $f$ is the outer function, which seems you have changed to be here.
    $endgroup$
    – xldd
    Dec 8 '18 at 10:30






  • 1




    $begingroup$
    It is not hard to alter the inner function. Try to do it yourself. Anyway I edited it for correctness.
    $endgroup$
    – Kemono Chen
    Dec 10 '18 at 2:45














2












2








2





$begingroup$

For question 1, $$f(x)=g(x)=
left{
begin{array}{ll}
x, &xin mathbb{Q} \
1-x,&xnotin mathbb{Q} \
end{array}
right.
\$$and $[a,b]=[0,1]$.



For question 2, see here.
In order to change the range of $f$, alter it to
$$ f(x) = begin{cases} 1/q & text{ for }x=p/qtext{ and $0le xle1/2$} \ 0 & text{ for } x notin mathbb{Q}text{ and $0le xle1/2$}\
2x-1 & text{others} end{cases}
$$ and the $g$ desired is the $f$ in my link.






share|cite|improve this answer











$endgroup$



For question 1, $$f(x)=g(x)=
left{
begin{array}{ll}
x, &xin mathbb{Q} \
1-x,&xnotin mathbb{Q} \
end{array}
right.
\$$and $[a,b]=[0,1]$.



For question 2, see here.
In order to change the range of $f$, alter it to
$$ f(x) = begin{cases} 1/q & text{ for }x=p/qtext{ and $0le xle1/2$} \ 0 & text{ for } x notin mathbb{Q}text{ and $0le xle1/2$}\
2x-1 & text{others} end{cases}
$$ and the $g$ desired is the $f$ in my link.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 2:44

























answered Dec 8 '18 at 5:25









Kemono ChenKemono Chen

3,1941844




3,1941844












  • $begingroup$
    question 2, there is another condition: the range of $f$ is $[a,b]$.
    $endgroup$
    – xldd
    Dec 8 '18 at 7:29










  • $begingroup$
    @xldd edited. Please check.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 7:53










  • $begingroup$
    @Chen What is the outer function and the inner function? It really troubles me.
    $endgroup$
    – xldd
    Dec 8 '18 at 10:28












  • $begingroup$
    In math.stackexchange.com/questions/1060834/…, the inner function is the Riemann function, whose range is not all of $[0,1]$, and the $f$ is the outer function, which seems you have changed to be here.
    $endgroup$
    – xldd
    Dec 8 '18 at 10:30






  • 1




    $begingroup$
    It is not hard to alter the inner function. Try to do it yourself. Anyway I edited it for correctness.
    $endgroup$
    – Kemono Chen
    Dec 10 '18 at 2:45


















  • $begingroup$
    question 2, there is another condition: the range of $f$ is $[a,b]$.
    $endgroup$
    – xldd
    Dec 8 '18 at 7:29










  • $begingroup$
    @xldd edited. Please check.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 7:53










  • $begingroup$
    @Chen What is the outer function and the inner function? It really troubles me.
    $endgroup$
    – xldd
    Dec 8 '18 at 10:28












  • $begingroup$
    In math.stackexchange.com/questions/1060834/…, the inner function is the Riemann function, whose range is not all of $[0,1]$, and the $f$ is the outer function, which seems you have changed to be here.
    $endgroup$
    – xldd
    Dec 8 '18 at 10:30






  • 1




    $begingroup$
    It is not hard to alter the inner function. Try to do it yourself. Anyway I edited it for correctness.
    $endgroup$
    – Kemono Chen
    Dec 10 '18 at 2:45
















$begingroup$
question 2, there is another condition: the range of $f$ is $[a,b]$.
$endgroup$
– xldd
Dec 8 '18 at 7:29




$begingroup$
question 2, there is another condition: the range of $f$ is $[a,b]$.
$endgroup$
– xldd
Dec 8 '18 at 7:29












$begingroup$
@xldd edited. Please check.
$endgroup$
– Kemono Chen
Dec 8 '18 at 7:53




$begingroup$
@xldd edited. Please check.
$endgroup$
– Kemono Chen
Dec 8 '18 at 7:53












$begingroup$
@Chen What is the outer function and the inner function? It really troubles me.
$endgroup$
– xldd
Dec 8 '18 at 10:28






$begingroup$
@Chen What is the outer function and the inner function? It really troubles me.
$endgroup$
– xldd
Dec 8 '18 at 10:28














$begingroup$
In math.stackexchange.com/questions/1060834/…, the inner function is the Riemann function, whose range is not all of $[0,1]$, and the $f$ is the outer function, which seems you have changed to be here.
$endgroup$
– xldd
Dec 8 '18 at 10:30




$begingroup$
In math.stackexchange.com/questions/1060834/…, the inner function is the Riemann function, whose range is not all of $[0,1]$, and the $f$ is the outer function, which seems you have changed to be here.
$endgroup$
– xldd
Dec 8 '18 at 10:30




1




1




$begingroup$
It is not hard to alter the inner function. Try to do it yourself. Anyway I edited it for correctness.
$endgroup$
– Kemono Chen
Dec 10 '18 at 2:45




$begingroup$
It is not hard to alter the inner function. Try to do it yourself. Anyway I edited it for correctness.
$endgroup$
– Kemono Chen
Dec 10 '18 at 2:45


















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