0

Intuitively why must the primitive $F$ of a function $f$ that is Riemann integrable in $[a,b]$, be continuous in $[a,b]$?

Thanks

themagiciant95
  • 161
  • Possible duplicate of http://math.stackexchange.com/questions/429769/is-an-integral-always-continuous –  Apr 03 '17 at 08:34
  • No , my question is what may happens is the primitive F is continuous in [a,b] and we assure that f is Riemann integrable – themagiciant95 Apr 03 '17 at 08:39
  • Because f is bounded hence F is Lipschitz. – Did Apr 03 '17 at 08:51
  • 3
    I think the answer to this question may depend on what one put into the word "primitive function". What is your definition? – mickep Apr 03 '17 at 10:29

1 Answers1

1

If $f$ is Riemann integrable on $[a,b]$ then there is an $M$ with $|f(t)|\leq M$ for all $t\in[a,b]$. Now let $F$ be a primitive of $f$ on $[a,b]$, for example $$F(x):=\int_a^x f(t)\>dt\qquad(a\leq x\leq b)\ .$$ Then for arbitrary $a\leq x<y\leq b$ one has $$F(y)-F(x)=\int_x^y f(t)\>dt\ ,$$ and therefore $$\bigl|F(y)-F(x)\bigr|\leq \int_x^y \bigl|f(t)\bigr|\>dt\leq M|y-x|\ .$$ This implies that $F$ is $M$-Lipschitz continuous on $[a,b]$.

Christian Blatter
  • 226,825
  • What justify the existence of $M$ upper bound of $|f|$ – Guy Fsone Jun 16 '17 at 19:14
  • I mean why is $|f| $ bounded? – Guy Fsone Jun 16 '17 at 19:15
  • @GuyFabrice: This is a basic requirement of Riemann integrability under any definition. When $f$ is unbounded then the set of possible Riemann sums will always be unbounded, however fine the chosen partition. – Christian Blatter Jun 16 '17 at 20:36
  • Yes your right but I am wondering why that fact does not apply to the function $x\mapsto |x|^{-a}$ with $0<a<1$ which is Riemann integrable on $(-1,1)$ but bounded. – Guy Fsone Jun 17 '17 at 10:47
  • @GuyFabrice: This function is not Riemann integrable over $[-1,1]$, but the improper Riemann integral $\lim_{\epsilon\to0+}\int_{[-1,-\epsilon]\cup[\epsilon,1]}\ldots$ exists. – Christian Blatter Jun 17 '17 at 12:06