If $f(x) geq g(x) forall x in [0,1]$, prove that $int_0^1 f geq int_0^1 g$












3












$begingroup$


Suppose that $f, g: [0,1] rightarrow mathbb{R}$ are Riemann integrable and $f(x) geq g(x) forall x in [0,1]$. Show that $int_0^1 f geq int_0^1 g$.



So this should be relatively simple. We have $f$ is Riemann integrable
if $L(f) = U(f)$. Where $L(f) = sup{L(f,P) | text{P a partition}}$, $U(f) = inf{U(f,P) | text{P a partition}}$.



Now we have that $f(x) geq g(x) forall x in [0,1]$. Because this holds, we have $L(f) geq L(g)$. Since $int_0^1$ is the common value $L(f) = U(f)$. We have $U(f) geq U(g)$ as well. And so $int_0^1 f geq int_0^1 g$.



Not too versed in proving stuff of this sort so would appreciate help/clarification.






real-analysis integration functions proof-verification riemann-integration







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I see the pieces of a complete proof... just oddly ordered. You have in the assumptions that $f$ and $g$ are integrable. Don't need to show it.
    $endgroup$
    – Robert Wolfe
    Dec 16 '18 at 5:16








  • 1




    $begingroup$
    I suppose you've glossed over an important piece. If you know $f(x)geq g(x)$ for all $x$, how do you know that $L(f,P)geq L(g,P)$ for any $P$ immediately? Lower/upper sums aren't built directly from functions values. (This would be easy with Riemann sums though)
    $endgroup$
    – Robert Wolfe
    Dec 16 '18 at 5:20










  • $begingroup$
    With doing it with Riemann sums, would it be correct to say that considering the Riemann sum, $I(f, P, {t_i}) = sum_{i=1}^n f(t_i) cdot (x_i - x_i-1)$, we have that this sum is larger for a partition with respect to $f$ than with $g$. So therefore, we have that the integral is greater.
    $endgroup$
    – SS'
    Dec 16 '18 at 5:26






  • 1




    $begingroup$
    Pretty much. With Riemann sums, the inequalities are almost immediate just from ordered field properties.
    $endgroup$
    – Robert Wolfe
    Dec 16 '18 at 5:29






  • 1




    $begingroup$
    It might be easier if you let $h(x)=f(x)-g(x)$, so $h(x) ge 0 , forall x in [0,1]$. Now show that $int_0^1 h(x) dx = int_0^1 f(x) dx-int_0^1 g(x) dx ge 0$.
    $endgroup$
    – tonychow0929
    Dec 16 '18 at 8:01


















3












$begingroup$


Suppose that $f, g: [0,1] rightarrow mathbb{R}$ are Riemann integrable and $f(x) geq g(x) forall x in [0,1]$. Show that $int_0^1 f geq int_0^1 g$.



So this should be relatively simple. We have $f$ is Riemann integrable
if $L(f) = U(f)$. Where $L(f) = sup{L(f,P) | text{P a partition}}$, $U(f) = inf{U(f,P) | text{P a partition}}$.



Now we have that $f(x) geq g(x) forall x in [0,1]$. Because this holds, we have $L(f) geq L(g)$. Since $int_0^1$ is the common value $L(f) = U(f)$. We have $U(f) geq U(g)$ as well. And so $int_0^1 f geq int_0^1 g$.



Not too versed in proving stuff of this sort so would appreciate help/clarification.






real-analysis integration functions proof-verification riemann-integration







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I see the pieces of a complete proof... just oddly ordered. You have in the assumptions that $f$ and $g$ are integrable. Don't need to show it.
    $endgroup$
    – Robert Wolfe
    Dec 16 '18 at 5:16








  • 1




    $begingroup$
    I suppose you've glossed over an important piece. If you know $f(x)geq g(x)$ for all $x$, how do you know that $L(f,P)geq L(g,P)$ for any $P$ immediately? Lower/upper sums aren't built directly from functions values. (This would be easy with Riemann sums though)
    $endgroup$
    – Robert Wolfe
    Dec 16 '18 at 5:20










  • $begingroup$
    With doing it with Riemann sums, would it be correct to say that considering the Riemann sum, $I(f, P, {t_i}) = sum_{i=1}^n f(t_i) cdot (x_i - x_i-1)$, we have that this sum is larger for a partition with respect to $f$ than with $g$. So therefore, we have that the integral is greater.
    $endgroup$
    – SS'
    Dec 16 '18 at 5:26






  • 1




    $begingroup$
    Pretty much. With Riemann sums, the inequalities are almost immediate just from ordered field properties.
    $endgroup$
    – Robert Wolfe
    Dec 16 '18 at 5:29






  • 1




    $begingroup$
    It might be easier if you let $h(x)=f(x)-g(x)$, so $h(x) ge 0 , forall x in [0,1]$. Now show that $int_0^1 h(x) dx = int_0^1 f(x) dx-int_0^1 g(x) dx ge 0$.
    $endgroup$
    – tonychow0929
    Dec 16 '18 at 8:01
















3












3








3





$begingroup$


Suppose that $f, g: [0,1] rightarrow mathbb{R}$ are Riemann integrable and $f(x) geq g(x) forall x in [0,1]$. Show that $int_0^1 f geq int_0^1 g$.



So this should be relatively simple. We have $f$ is Riemann integrable
if $L(f) = U(f)$. Where $L(f) = sup{L(f,P) | text{P a partition}}$, $U(f) = inf{U(f,P) | text{P a partition}}$.



Now we have that $f(x) geq g(x) forall x in [0,1]$. Because this holds, we have $L(f) geq L(g)$. Since $int_0^1$ is the common value $L(f) = U(f)$. We have $U(f) geq U(g)$ as well. And so $int_0^1 f geq int_0^1 g$.



Not too versed in proving stuff of this sort so would appreciate help/clarification.






real-analysis integration functions proof-verification riemann-integration







share|cite|improve this question









$endgroup$




Suppose that $f, g: [0,1] rightarrow mathbb{R}$ are Riemann integrable and $f(x) geq g(x) forall x in [0,1]$. Show that $int_0^1 f geq int_0^1 g$.



So this should be relatively simple. We have $f$ is Riemann integrable
if $L(f) = U(f)$. Where $L(f) = sup{L(f,P) | text{P a partition}}$, $U(f) = inf{U(f,P) | text{P a partition}}$.



Now we have that $f(x) geq g(x) forall x in [0,1]$. Because this holds, we have $L(f) geq L(g)$. Since $int_0^1$ is the common value $L(f) = U(f)$. We have $U(f) geq U(g)$ as well. And so $int_0^1 f geq int_0^1 g$.



Not too versed in proving stuff of this sort so would appreciate help/clarification.






real-analysis integration functions proof-verification riemann-integration




real-analysis integration functions proof-verification riemann-integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 16 '18 at 5:07









SS'SS'

592314




592314








  • 1




    $begingroup$
    I see the pieces of a complete proof... just oddly ordered. You have in the assumptions that $f$ and $g$ are integrable. Don't need to show it.
    $endgroup$
    – Robert Wolfe
    Dec 16 '18 at 5:16








  • 1




    $begingroup$
    I suppose you've glossed over an important piece. If you know $f(x)geq g(x)$ for all $x$, how do you know that $L(f,P)geq L(g,P)$ for any $P$ immediately? Lower/upper sums aren't built directly from functions values. (This would be easy with Riemann sums though)
    $endgroup$
    – Robert Wolfe
    Dec 16 '18 at 5:20










  • $begingroup$
    With doing it with Riemann sums, would it be correct to say that considering the Riemann sum, $I(f, P, {t_i}) = sum_{i=1}^n f(t_i) cdot (x_i - x_i-1)$, we have that this sum is larger for a partition with respect to $f$ than with $g$. So therefore, we have that the integral is greater.
    $endgroup$
    – SS'
    Dec 16 '18 at 5:26






  • 1




    $begingroup$
    Pretty much. With Riemann sums, the inequalities are almost immediate just from ordered field properties.
    $endgroup$
    – Robert Wolfe
    Dec 16 '18 at 5:29






  • 1




    $begingroup$
    It might be easier if you let $h(x)=f(x)-g(x)$, so $h(x) ge 0 , forall x in [0,1]$. Now show that $int_0^1 h(x) dx = int_0^1 f(x) dx-int_0^1 g(x) dx ge 0$.
    $endgroup$
    – tonychow0929
    Dec 16 '18 at 8:01
















  • 1




    $begingroup$
    I see the pieces of a complete proof... just oddly ordered. You have in the assumptions that $f$ and $g$ are integrable. Don't need to show it.
    $endgroup$
    – Robert Wolfe
    Dec 16 '18 at 5:16








  • 1




    $begingroup$
    I suppose you've glossed over an important piece. If you know $f(x)geq g(x)$ for all $x$, how do you know that $L(f,P)geq L(g,P)$ for any $P$ immediately? Lower/upper sums aren't built directly from functions values. (This would be easy with Riemann sums though)
    $endgroup$
    – Robert Wolfe
    Dec 16 '18 at 5:20










  • $begingroup$
    With doing it with Riemann sums, would it be correct to say that considering the Riemann sum, $I(f, P, {t_i}) = sum_{i=1}^n f(t_i) cdot (x_i - x_i-1)$, we have that this sum is larger for a partition with respect to $f$ than with $g$. So therefore, we have that the integral is greater.
    $endgroup$
    – SS'
    Dec 16 '18 at 5:26






  • 1




    $begingroup$
    Pretty much. With Riemann sums, the inequalities are almost immediate just from ordered field properties.
    $endgroup$
    – Robert Wolfe
    Dec 16 '18 at 5:29






  • 1




    $begingroup$
    It might be easier if you let $h(x)=f(x)-g(x)$, so $h(x) ge 0 , forall x in [0,1]$. Now show that $int_0^1 h(x) dx = int_0^1 f(x) dx-int_0^1 g(x) dx ge 0$.
    $endgroup$
    – tonychow0929
    Dec 16 '18 at 8:01










1




1




$begingroup$
I see the pieces of a complete proof... just oddly ordered. You have in the assumptions that $f$ and $g$ are integrable. Don't need to show it.
$endgroup$
– Robert Wolfe
Dec 16 '18 at 5:16






$begingroup$
I see the pieces of a complete proof... just oddly ordered. You have in the assumptions that $f$ and $g$ are integrable. Don't need to show it.
$endgroup$
– Robert Wolfe
Dec 16 '18 at 5:16






1




1




$begingroup$
I suppose you've glossed over an important piece. If you know $f(x)geq g(x)$ for all $x$, how do you know that $L(f,P)geq L(g,P)$ for any $P$ immediately? Lower/upper sums aren't built directly from functions values. (This would be easy with Riemann sums though)
$endgroup$
– Robert Wolfe
Dec 16 '18 at 5:20




$begingroup$
I suppose you've glossed over an important piece. If you know $f(x)geq g(x)$ for all $x$, how do you know that $L(f,P)geq L(g,P)$ for any $P$ immediately? Lower/upper sums aren't built directly from functions values. (This would be easy with Riemann sums though)
$endgroup$
– Robert Wolfe
Dec 16 '18 at 5:20












$begingroup$
With doing it with Riemann sums, would it be correct to say that considering the Riemann sum, $I(f, P, {t_i}) = sum_{i=1}^n f(t_i) cdot (x_i - x_i-1)$, we have that this sum is larger for a partition with respect to $f$ than with $g$. So therefore, we have that the integral is greater.
$endgroup$
– SS'
Dec 16 '18 at 5:26




$begingroup$
With doing it with Riemann sums, would it be correct to say that considering the Riemann sum, $I(f, P, {t_i}) = sum_{i=1}^n f(t_i) cdot (x_i - x_i-1)$, we have that this sum is larger for a partition with respect to $f$ than with $g$. So therefore, we have that the integral is greater.
$endgroup$
– SS'
Dec 16 '18 at 5:26




1




1




$begingroup$
Pretty much. With Riemann sums, the inequalities are almost immediate just from ordered field properties.
$endgroup$
– Robert Wolfe
Dec 16 '18 at 5:29




$begingroup$
Pretty much. With Riemann sums, the inequalities are almost immediate just from ordered field properties.
$endgroup$
– Robert Wolfe
Dec 16 '18 at 5:29




1




1




$begingroup$
It might be easier if you let $h(x)=f(x)-g(x)$, so $h(x) ge 0 , forall x in [0,1]$. Now show that $int_0^1 h(x) dx = int_0^1 f(x) dx-int_0^1 g(x) dx ge 0$.
$endgroup$
– tonychow0929
Dec 16 '18 at 8:01






$begingroup$
It might be easier if you let $h(x)=f(x)-g(x)$, so $h(x) ge 0 , forall x in [0,1]$. Now show that $int_0^1 h(x) dx = int_0^1 f(x) dx-int_0^1 g(x) dx ge 0$.
$endgroup$
– tonychow0929
Dec 16 '18 at 8:01












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$begingroup$

There's nothing wrong with it, but for someone at your stage, I would expect some explanation of why $f(x) ge g(x)$ implies $L(f) ge L(g)$, especially since this is the main part of the proof.






share|cite|improve this answer









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    active

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    2












    $begingroup$

    There's nothing wrong with it, but for someone at your stage, I would expect some explanation of why $f(x) ge g(x)$ implies $L(f) ge L(g)$, especially since this is the main part of the proof.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      There's nothing wrong with it, but for someone at your stage, I would expect some explanation of why $f(x) ge g(x)$ implies $L(f) ge L(g)$, especially since this is the main part of the proof.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        There's nothing wrong with it, but for someone at your stage, I would expect some explanation of why $f(x) ge g(x)$ implies $L(f) ge L(g)$, especially since this is the main part of the proof.






        share|cite|improve this answer









        $endgroup$



        There's nothing wrong with it, but for someone at your stage, I would expect some explanation of why $f(x) ge g(x)$ implies $L(f) ge L(g)$, especially since this is the main part of the proof.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 5:18









        Trevor GunnTrevor Gunn

        14.8k32047




        14.8k32047






























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