A question regarding Riemann Integrability












0












$begingroup$


Let's say that $f : [a,b] → ℝ$ is such that:




  • $f$ is Riemann integrable on $[a,x]$ for each $x in [a,b)$


  • $f$ is continuous in $b$ (but not necessarily in $[a,b]$)



Is it true that $f$ is Riemann integrable on $[a,b]$?



And if we add the hypothesis $lim_{x→b} ∫_a^xf(t)dt ∈ ℝ$?






integration







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    If you know for all $x in [a,b]$ that $f$ is Riemann integrable on $[a,x]$, and for every $delta>0$ $f$ is bounded on $(b-delta, b]$, this is enough to conclude Riemann integrability on $[a,b]$.
    $endgroup$
    – MathematicsStudent1122
    Dec 25 '18 at 10:53












  • $begingroup$
    @MathematicsStudent1122 Why do you say that if $f$ it's bounded on $(b-delta,b]$ then it is integrable?
    $endgroup$
    – Markus Steiner
    Dec 25 '18 at 10:58












  • $begingroup$
    @MarkusSteiner because you have continuity at $b$ which implies existence of the limit which implies boundedness in a neighborhood.
    $endgroup$
    – John11
    Dec 25 '18 at 11:09










  • $begingroup$
    @John11 he asked why bounded implies integrable, not why bounded.
    $endgroup$
    – mathworker21
    Dec 25 '18 at 11:48










  • $begingroup$
    @mathworker21 the comment was edited.
    $endgroup$
    – John11
    Dec 25 '18 at 12:43
















0












$begingroup$


Let's say that $f : [a,b] → ℝ$ is such that:




  • $f$ is Riemann integrable on $[a,x]$ for each $x in [a,b)$


  • $f$ is continuous in $b$ (but not necessarily in $[a,b]$)



Is it true that $f$ is Riemann integrable on $[a,b]$?



And if we add the hypothesis $lim_{x→b} ∫_a^xf(t)dt ∈ ℝ$?






integration







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    If you know for all $x in [a,b]$ that $f$ is Riemann integrable on $[a,x]$, and for every $delta>0$ $f$ is bounded on $(b-delta, b]$, this is enough to conclude Riemann integrability on $[a,b]$.
    $endgroup$
    – MathematicsStudent1122
    Dec 25 '18 at 10:53












  • $begingroup$
    @MathematicsStudent1122 Why do you say that if $f$ it's bounded on $(b-delta,b]$ then it is integrable?
    $endgroup$
    – Markus Steiner
    Dec 25 '18 at 10:58












  • $begingroup$
    @MarkusSteiner because you have continuity at $b$ which implies existence of the limit which implies boundedness in a neighborhood.
    $endgroup$
    – John11
    Dec 25 '18 at 11:09










  • $begingroup$
    @John11 he asked why bounded implies integrable, not why bounded.
    $endgroup$
    – mathworker21
    Dec 25 '18 at 11:48










  • $begingroup$
    @mathworker21 the comment was edited.
    $endgroup$
    – John11
    Dec 25 '18 at 12:43














0












0








0


1



$begingroup$


Let's say that $f : [a,b] → ℝ$ is such that:




  • $f$ is Riemann integrable on $[a,x]$ for each $x in [a,b)$


  • $f$ is continuous in $b$ (but not necessarily in $[a,b]$)



Is it true that $f$ is Riemann integrable on $[a,b]$?



And if we add the hypothesis $lim_{x→b} ∫_a^xf(t)dt ∈ ℝ$?






integration







share|cite|improve this question









$endgroup$




Let's say that $f : [a,b] → ℝ$ is such that:




  • $f$ is Riemann integrable on $[a,x]$ for each $x in [a,b)$


  • $f$ is continuous in $b$ (but not necessarily in $[a,b]$)



Is it true that $f$ is Riemann integrable on $[a,b]$?



And if we add the hypothesis $lim_{x→b} ∫_a^xf(t)dt ∈ ℝ$?






integration




integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 25 '18 at 10:47









Markus SteinerMarkus Steiner

1036




1036








  • 2




    $begingroup$
    If you know for all $x in [a,b]$ that $f$ is Riemann integrable on $[a,x]$, and for every $delta>0$ $f$ is bounded on $(b-delta, b]$, this is enough to conclude Riemann integrability on $[a,b]$.
    $endgroup$
    – MathematicsStudent1122
    Dec 25 '18 at 10:53












  • $begingroup$
    @MathematicsStudent1122 Why do you say that if $f$ it's bounded on $(b-delta,b]$ then it is integrable?
    $endgroup$
    – Markus Steiner
    Dec 25 '18 at 10:58












  • $begingroup$
    @MarkusSteiner because you have continuity at $b$ which implies existence of the limit which implies boundedness in a neighborhood.
    $endgroup$
    – John11
    Dec 25 '18 at 11:09










  • $begingroup$
    @John11 he asked why bounded implies integrable, not why bounded.
    $endgroup$
    – mathworker21
    Dec 25 '18 at 11:48










  • $begingroup$
    @mathworker21 the comment was edited.
    $endgroup$
    – John11
    Dec 25 '18 at 12:43














  • 2




    $begingroup$
    If you know for all $x in [a,b]$ that $f$ is Riemann integrable on $[a,x]$, and for every $delta>0$ $f$ is bounded on $(b-delta, b]$, this is enough to conclude Riemann integrability on $[a,b]$.
    $endgroup$
    – MathematicsStudent1122
    Dec 25 '18 at 10:53












  • $begingroup$
    @MathematicsStudent1122 Why do you say that if $f$ it's bounded on $(b-delta,b]$ then it is integrable?
    $endgroup$
    – Markus Steiner
    Dec 25 '18 at 10:58












  • $begingroup$
    @MarkusSteiner because you have continuity at $b$ which implies existence of the limit which implies boundedness in a neighborhood.
    $endgroup$
    – John11
    Dec 25 '18 at 11:09










  • $begingroup$
    @John11 he asked why bounded implies integrable, not why bounded.
    $endgroup$
    – mathworker21
    Dec 25 '18 at 11:48










  • $begingroup$
    @mathworker21 the comment was edited.
    $endgroup$
    – John11
    Dec 25 '18 at 12:43








2




2




$begingroup$
If you know for all $x in [a,b]$ that $f$ is Riemann integrable on $[a,x]$, and for every $delta>0$ $f$ is bounded on $(b-delta, b]$, this is enough to conclude Riemann integrability on $[a,b]$.
$endgroup$
– MathematicsStudent1122
Dec 25 '18 at 10:53






$begingroup$
If you know for all $x in [a,b]$ that $f$ is Riemann integrable on $[a,x]$, and for every $delta>0$ $f$ is bounded on $(b-delta, b]$, this is enough to conclude Riemann integrability on $[a,b]$.
$endgroup$
– MathematicsStudent1122
Dec 25 '18 at 10:53














$begingroup$
@MathematicsStudent1122 Why do you say that if $f$ it's bounded on $(b-delta,b]$ then it is integrable?
$endgroup$
– Markus Steiner
Dec 25 '18 at 10:58






$begingroup$
@MathematicsStudent1122 Why do you say that if $f$ it's bounded on $(b-delta,b]$ then it is integrable?
$endgroup$
– Markus Steiner
Dec 25 '18 at 10:58














$begingroup$
@MarkusSteiner because you have continuity at $b$ which implies existence of the limit which implies boundedness in a neighborhood.
$endgroup$
– John11
Dec 25 '18 at 11:09




$begingroup$
@MarkusSteiner because you have continuity at $b$ which implies existence of the limit which implies boundedness in a neighborhood.
$endgroup$
– John11
Dec 25 '18 at 11:09












$begingroup$
@John11 he asked why bounded implies integrable, not why bounded.
$endgroup$
– mathworker21
Dec 25 '18 at 11:48




$begingroup$
@John11 he asked why bounded implies integrable, not why bounded.
$endgroup$
– mathworker21
Dec 25 '18 at 11:48












$begingroup$
@mathworker21 the comment was edited.
$endgroup$
– John11
Dec 25 '18 at 12:43




$begingroup$
@mathworker21 the comment was edited.
$endgroup$
– John11
Dec 25 '18 at 12:43










1 Answer
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$begingroup$

If $f$ is bounded in magnitude (say by $M$) on $[a,b]$ and Riemann integrable on $[a,x]$ for all $x in [a,b]$, then it is Riemann integrable on $[a,b]$.



Let $epsilon>0$ and put $x = text{min}left(frac{epsilon}{4M}, frac{b-a}{2}right)$. Choose a partition $P$ of $[a, b-x]$ so that $U_P - L_P < frac{epsilon}{2}$. Our partition $P'$ of $[a,b]$ will simply be $P' = P cup {b}$ and observe that $U_{P'} - L_{P'} = U_{P} - L_{P} + x(sup_{x in [b-x, b]} f(x) - inf_{x in [b-x,b]} f(x)) leq U_p - L_p + frac{epsilon}{4M} 2M < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.



Since proper Riemann integrability always presupposes boundedness, this more or less always holds.






share|cite|improve this answer









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    $begingroup$

    If $f$ is bounded in magnitude (say by $M$) on $[a,b]$ and Riemann integrable on $[a,x]$ for all $x in [a,b]$, then it is Riemann integrable on $[a,b]$.



    Let $epsilon>0$ and put $x = text{min}left(frac{epsilon}{4M}, frac{b-a}{2}right)$. Choose a partition $P$ of $[a, b-x]$ so that $U_P - L_P < frac{epsilon}{2}$. Our partition $P'$ of $[a,b]$ will simply be $P' = P cup {b}$ and observe that $U_{P'} - L_{P'} = U_{P} - L_{P} + x(sup_{x in [b-x, b]} f(x) - inf_{x in [b-x,b]} f(x)) leq U_p - L_p + frac{epsilon}{4M} 2M < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.



    Since proper Riemann integrability always presupposes boundedness, this more or less always holds.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      If $f$ is bounded in magnitude (say by $M$) on $[a,b]$ and Riemann integrable on $[a,x]$ for all $x in [a,b]$, then it is Riemann integrable on $[a,b]$.



      Let $epsilon>0$ and put $x = text{min}left(frac{epsilon}{4M}, frac{b-a}{2}right)$. Choose a partition $P$ of $[a, b-x]$ so that $U_P - L_P < frac{epsilon}{2}$. Our partition $P'$ of $[a,b]$ will simply be $P' = P cup {b}$ and observe that $U_{P'} - L_{P'} = U_{P} - L_{P} + x(sup_{x in [b-x, b]} f(x) - inf_{x in [b-x,b]} f(x)) leq U_p - L_p + frac{epsilon}{4M} 2M < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.



      Since proper Riemann integrability always presupposes boundedness, this more or less always holds.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        If $f$ is bounded in magnitude (say by $M$) on $[a,b]$ and Riemann integrable on $[a,x]$ for all $x in [a,b]$, then it is Riemann integrable on $[a,b]$.



        Let $epsilon>0$ and put $x = text{min}left(frac{epsilon}{4M}, frac{b-a}{2}right)$. Choose a partition $P$ of $[a, b-x]$ so that $U_P - L_P < frac{epsilon}{2}$. Our partition $P'$ of $[a,b]$ will simply be $P' = P cup {b}$ and observe that $U_{P'} - L_{P'} = U_{P} - L_{P} + x(sup_{x in [b-x, b]} f(x) - inf_{x in [b-x,b]} f(x)) leq U_p - L_p + frac{epsilon}{4M} 2M < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.



        Since proper Riemann integrability always presupposes boundedness, this more or less always holds.






        share|cite|improve this answer









        $endgroup$



        If $f$ is bounded in magnitude (say by $M$) on $[a,b]$ and Riemann integrable on $[a,x]$ for all $x in [a,b]$, then it is Riemann integrable on $[a,b]$.



        Let $epsilon>0$ and put $x = text{min}left(frac{epsilon}{4M}, frac{b-a}{2}right)$. Choose a partition $P$ of $[a, b-x]$ so that $U_P - L_P < frac{epsilon}{2}$. Our partition $P'$ of $[a,b]$ will simply be $P' = P cup {b}$ and observe that $U_{P'} - L_{P'} = U_{P} - L_{P} + x(sup_{x in [b-x, b]} f(x) - inf_{x in [b-x,b]} f(x)) leq U_p - L_p + frac{epsilon}{4M} 2M < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$.



        Since proper Riemann integrability always presupposes boundedness, this more or less always holds.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 11:07









        MathematicsStudent1122MathematicsStudent1122

        9,01432669




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