Examples of real-valued functions $f,g$ such that $L(f+g, [0,1])neq L(f, [0,1])+L(g, [0,1])$ and $U(f+g, [0,1])neq U(f, [0,1])+L(g, [0,1])$?
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As per the title, I need to come up with real-valued functions $f$ and $g$ such that $$L(f+g, [0,1])neq L(f, [0,1])+L(g, [0,1])$$ and $$U(f+g, [0,1])neq U(f, [0,1])+L(g, [0,1]).$$ So in other words, I need to find $f$ and $g$ such that
$$underlineint_0^1 (f+g)neq underlineint_0^1 f+underlineint_0^1 g$$ and $$overlineint_0^1 (f+g)neq overlineint_0^1 f+overlineint_0^1 g.$$ I feel like the only functions $f$ and $g$ that would work here are ones that are not Riemann integrable. I know that a classic example of a non-Riemann integrable function is the Dirichlet function, i.e. $$f(x)=begincases 1 & x text is rational \
0 & xtext is irrationalendcases\$$ which has $L(f,[0,1])=0$ and $U(f,[0,1])=1$. But (assuming I am on the right track here) I still need another function $g$ and somehow have to consider $f+g$.
Any suggestions with this one?
real-analysis
asked Sep 6 at 4:54
Thy Art is Math
37819
 |Â
show 2 more comments
up vote
0
down vote
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As per the title, I need to come up with real-valued functions $f$ and $g$ such that $$L(f+g, [0,1])neq L(f, [0,1])+L(g, [0,1])$$ and $$U(f+g, [0,1])neq U(f, [0,1])+L(g, [0,1]).$$ So in other words, I need to find $f$ and $g$ such that
$$underlineint_0^1 (f+g)neq underlineint_0^1 f+underlineint_0^1 g$$ and $$overlineint_0^1 (f+g)neq overlineint_0^1 f+overlineint_0^1 g.$$ I feel like the only functions $f$ and $g$ that would work here are ones that are not Riemann integrable. I know that a classic example of a non-Riemann integrable function is the Dirichlet function, i.e. $$f(x)=begincases 1 & x text is rational \
0 & xtext is irrationalendcases\$$ which has $L(f,[0,1])=0$ and $U(f,[0,1])=1$. But (assuming I am on the right track here) I still need another function $g$ and somehow have to consider $f+g$.
Any suggestions with this one?
real-analysis
asked Sep 6 at 4:54
Thy Art is Math
37819
1
Easy. Let $g = -f$.
– xbh
Sep 6 at 5:00
So would something like $f=cos(x), g=-cos(x)$ work?
– Thy Art is Math
Sep 6 at 5:00
1
Seems no, $f,g $ are Riemann integrable.
– xbh
Sep 6 at 5:02
1
Yes, $f$ is the Dirichlet function. Sorry I was ambiguous.
– xbh
Sep 6 at 5:03
1
@xbh you should post this as an answer(and not a comment)
– Holo
Sep 6 at 5:12
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
As per the title, I need to come up with real-valued functions $f$ and $g$ such that $$L(f+g, [0,1])neq L(f, [0,1])+L(g, [0,1])$$ and $$U(f+g, [0,1])neq U(f, [0,1])+L(g, [0,1]).$$ So in other words, I need to find $f$ and $g$ such that
$$underlineint_0^1 (f+g)neq underlineint_0^1 f+underlineint_0^1 g$$ and $$overlineint_0^1 (f+g)neq overlineint_0^1 f+overlineint_0^1 g.$$ I feel like the only functions $f$ and $g$ that would work here are ones that are not Riemann integrable. I know that a classic example of a non-Riemann integrable function is the Dirichlet function, i.e. $$f(x)=begincases 1 & x text is rational \
0 & xtext is irrationalendcases\$$ which has $L(f,[0,1])=0$ and $U(f,[0,1])=1$. But (assuming I am on the right track here) I still need another function $g$ and somehow have to consider $f+g$.
Any suggestions with this one?
real-analysis
asked Sep 6 at 4:54
Thy Art is Math
37819
As per the title, I need to come up with real-valued functions $f$ and $g$ such that $$L(f+g, [0,1])neq L(f, [0,1])+L(g, [0,1])$$ and $$U(f+g, [0,1])neq U(f, [0,1])+L(g, [0,1]).$$ So in other words, I need to find $f$ and $g$ such that
$$underlineint_0^1 (f+g)neq underlineint_0^1 f+underlineint_0^1 g$$ and $$overlineint_0^1 (f+g)neq overlineint_0^1 f+overlineint_0^1 g.$$ I feel like the only functions $f$ and $g$ that would work here are ones that are not Riemann integrable. I know that a classic example of a non-Riemann integrable function is the Dirichlet function, i.e. $$f(x)=begincases 1 & x text is rational \
0 & xtext is irrationalendcases\$$ which has $L(f,[0,1])=0$ and $U(f,[0,1])=1$. But (assuming I am on the right track here) I still need another function $g$ and somehow have to consider $f+g$.
Any suggestions with this one?
real-analysis
real-analysis
asked Sep 6 at 4:54
Thy Art is Math
37819
asked Sep 6 at 4:54
Thy Art is Math
37819
asked Sep 6 at 4:54
Thy Art is Math
37819
asked Sep 6 at 4:54
Thy Art is Math
37819
asked Sep 6 at 4:54
Thy Art is Math
37819
37819
1
Easy. Let $g = -f$.
– xbh
Sep 6 at 5:00
So would something like $f=cos(x), g=-cos(x)$ work?
– Thy Art is Math
Sep 6 at 5:00
1
Seems no, $f,g $ are Riemann integrable.
– xbh
Sep 6 at 5:02
1
Yes, $f$ is the Dirichlet function. Sorry I was ambiguous.
– xbh
Sep 6 at 5:03
1
@xbh you should post this as an answer(and not a comment)
– Holo
Sep 6 at 5:12
 |Â
show 2 more comments
1
Easy. Let $g = -f$.
– xbh
Sep 6 at 5:00
So would something like $f=cos(x), g=-cos(x)$ work?
– Thy Art is Math
Sep 6 at 5:00
1
Seems no, $f,g $ are Riemann integrable.
– xbh
Sep 6 at 5:02
1
Yes, $f$ is the Dirichlet function. Sorry I was ambiguous.
– xbh
Sep 6 at 5:03
1
@xbh you should post this as an answer(and not a comment)
– Holo
Sep 6 at 5:12
1
1
Easy. Let $g = -f$.
– xbh
Sep 6 at 5:00
Easy. Let $g = -f$.
– xbh
Sep 6 at 5:00
So would something like $f=cos(x), g=-cos(x)$ work?
– Thy Art is Math
Sep 6 at 5:00
So would something like $f=cos(x), g=-cos(x)$ work?
– Thy Art is Math
Sep 6 at 5:00
1
1
Seems no, $f,g $ are Riemann integrable.
– xbh
Sep 6 at 5:02
Seems no, $f,g $ are Riemann integrable.
– xbh
Sep 6 at 5:02
1
1
Yes, $f$ is the Dirichlet function. Sorry I was ambiguous.
– xbh
Sep 6 at 5:03
Yes, $f$ is the Dirichlet function. Sorry I was ambiguous.
– xbh
Sep 6 at 5:03
1
1
@xbh you should post this as an answer(and not a comment)
– Holo
Sep 6 at 5:12
@xbh you should post this as an answer(and not a comment)
– Holo
Sep 6 at 5:12
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Take $f = mathbb 1_mathbb Q cap [0,1]$, $g = -f$, then $f+g =0$ is always Riemann integrable, hence $L(f+g) = U(f+g) = 0$. But $U(f) = 1, L(f) = U(g) = 0, L(g)=-1$.
answered Sep 6 at 5:17
xbh
3,650320
Thanks for making this into an answer. What does $1_mathbbQcap[0,1]$ mean?
– Thy Art is Math
Sep 6 at 5:18
That's a standard notation of Dirichlet function. You could see indicator function in wiki.
– xbh
Sep 6 at 5:20
I was unaware of that. Thank you.
– Thy Art is Math
Sep 6 at 5:22
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Take $f = mathbb 1_mathbb Q cap [0,1]$, $g = -f$, then $f+g =0$ is always Riemann integrable, hence $L(f+g) = U(f+g) = 0$. But $U(f) = 1, L(f) = U(g) = 0, L(g)=-1$.
answered Sep 6 at 5:17
xbh
3,650320
Thanks for making this into an answer. What does $1_mathbbQcap[0,1]$ mean?
– Thy Art is Math
Sep 6 at 5:18
That's a standard notation of Dirichlet function. You could see indicator function in wiki.
– xbh
Sep 6 at 5:20
I was unaware of that. Thank you.
– Thy Art is Math
Sep 6 at 5:22
add a comment |Â
up vote
1
down vote
accepted
Take $f = mathbb 1_mathbb Q cap [0,1]$, $g = -f$, then $f+g =0$ is always Riemann integrable, hence $L(f+g) = U(f+g) = 0$. But $U(f) = 1, L(f) = U(g) = 0, L(g)=-1$.
answered Sep 6 at 5:17
xbh
3,650320
Thanks for making this into an answer. What does $1_mathbbQcap[0,1]$ mean?
– Thy Art is Math
Sep 6 at 5:18
That's a standard notation of Dirichlet function. You could see indicator function in wiki.
– xbh
Sep 6 at 5:20
I was unaware of that. Thank you.
– Thy Art is Math
Sep 6 at 5:22
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Take $f = mathbb 1_mathbb Q cap [0,1]$, $g = -f$, then $f+g =0$ is always Riemann integrable, hence $L(f+g) = U(f+g) = 0$. But $U(f) = 1, L(f) = U(g) = 0, L(g)=-1$.
answered Sep 6 at 5:17
xbh
3,650320
Take $f = mathbb 1_mathbb Q cap [0,1]$, $g = -f$, then $f+g =0$ is always Riemann integrable, hence $L(f+g) = U(f+g) = 0$. But $U(f) = 1, L(f) = U(g) = 0, L(g)=-1$.
answered Sep 6 at 5:17
xbh
3,650320
answered Sep 6 at 5:17
xbh
3,650320
answered Sep 6 at 5:17
xbh
3,650320
answered Sep 6 at 5:17
xbh
3,650320
3,650320
Thanks for making this into an answer. What does $1_mathbbQcap[0,1]$ mean?
– Thy Art is Math
Sep 6 at 5:18
That's a standard notation of Dirichlet function. You could see indicator function in wiki.
– xbh
Sep 6 at 5:20
I was unaware of that. Thank you.
– Thy Art is Math
Sep 6 at 5:22
add a comment |Â
Thanks for making this into an answer. What does $1_mathbbQcap[0,1]$ mean?
– Thy Art is Math
Sep 6 at 5:18
That's a standard notation of Dirichlet function. You could see indicator function in wiki.
– xbh
Sep 6 at 5:20
I was unaware of that. Thank you.
– Thy Art is Math
Sep 6 at 5:22
Thanks for making this into an answer. What does $1_mathbbQcap[0,1]$ mean?
– Thy Art is Math
Sep 6 at 5:18
Thanks for making this into an answer. What does $1_mathbbQcap[0,1]$ mean?
– Thy Art is Math
Sep 6 at 5:18
That's a standard notation of Dirichlet function. You could see indicator function in wiki.
– xbh
Sep 6 at 5:20
That's a standard notation of Dirichlet function. You could see indicator function in wiki.
– xbh
Sep 6 at 5:20
I was unaware of that. Thank you.
– Thy Art is Math
Sep 6 at 5:22
I was unaware of that. Thank you.
– Thy Art is Math
Sep 6 at 5:22
add a comment |Â
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1
Easy. Let $g = -f$.
– xbh
Sep 6 at 5:00
So would something like $f=cos(x), g=-cos(x)$ work?
– Thy Art is Math
Sep 6 at 5:00
1
Seems no, $f,g $ are Riemann integrable.
– xbh
Sep 6 at 5:02
1
Yes, $f$ is the Dirichlet function. Sorry I was ambiguous.
– xbh
Sep 6 at 5:03
1
@xbh you should post this as an answer(and not a comment)
– Holo
Sep 6 at 5:12