0
$\begingroup$

Let $G(x)=\int_a^xg(t)dt$ for some $g(t):[a,b]\rightarrow \mathbb{R}$ ($g(t)$ is Riemann integrable on $[a,b]$) , and suppose we know that $G(x)$ is differentiable on $[a,b]$. Does it follow that $G'(x)$ is Riemann integrable on $[a,b]$?

I believe this is true, but I'm completely at a loss on how to prove it, and would really appreciate any hints.

Edit: I would like to know if this is true even if $g$ is discontinuous (but still integrable). The continuous case is clear from the fundamental theorem of calculs

$\endgroup$
1
  • $\begingroup$ @SimonS I don't want to restrict $g$ to be continuous, because that case is clear from the Fundamental Theorem of Calculus (Part II). I don't think you can use FTC when $g$ is discontinuous $\endgroup$
    – A Nonny
    Nov 11, 2014 at 2:31

1 Answer 1

1
$\begingroup$

Your suspect is well-founded: in fact $G'$ is continuous where $g$ is continuous.

Then, since $g$ is continuous a.e. on $[a,b]$, so also $G'$ is $\dots$

Let $x_0$ be a point of continuity for $g\,$: why is $G'$ continuous in $x_0$ ?

Informally, the reason is that, if $x$ is close to $x_0$ and $h$ is small, then $\frac {G(x+h)-G(x)}h$ is close to $g(x_0)$.

The $\varepsilon-\delta$ argument starts from $$|G'(x)-g(x_0)| \le \left|G'(x)- \frac {G(x+h)-G(x)}h \right| + \left|\frac {G(x+h)-G(x)}h - g(x_0) \right|$$ knowing that $$\frac {G(x+h)-G(x)}h - g(x_0)=\frac1h \int_x^{x+h} (g(t)-g(x_0))\,dt$$Let me know if you can't conclude.

$\endgroup$
1
  • $\begingroup$ Thanks! I proved it using Darboux criterion for integrability, but didn't realise that in fact $G'$ is continuous where $g$ is, so I really appreciate your answer! $\endgroup$
    – A Nonny
    Nov 25, 2014 at 13:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .