Let $f(x)$ be a Riemann integrable function (not neccesarily continuous) defined on some interval I. Let $F(x) = \int_a^xf(t)dt$ be it's integral function. Suppose that $F(x)$ is differentiable in the interval where it is defined, and let $F'(x)$ be its derivative. Is it true that $F'(x)$ is Riemann integrable on the interval I?
I was wondering if this proposition is true, since the symmetric proposition is true by Barrow's Rule (FTC)
Let $F(x)$ be a differentiable function defined on some interval I. Let $f(x) = F'(x)$ be it's derivative. Suppose that $f(x)$ is integrable in the interval where it is defined. Then $\int_a^x f(t)dt = F(x) + c$, where $c = -F(a)$
If the first proposition its true, applying the second one we would have $\int_a^x F'(t)dt = F(x) – F(a)$
I would be grateful if the the proof is "elementary" and does not use any Measure Theory or Lebesgue integral results.
Best Answer
We let $I=[a,b]$ and consider a partition $P=\{a=x_{0}<x_{1}<\ldots<x_{n}=b\}$ of $I$. Next observe that for each $s,t\in[x_{i-1},x_{i}]$, where $i=1,2,\ldots,n$ and $s<t$, that
\begin{equation*} \frac{F(t)-F(s)}{t-s}=\frac{1}{t-s}\int_{s}^{t}\!{}f(x)\mathrm{d}x \ge\inf_{[x_{i-1},x_{i}]}(f). \end{equation*}
Letting $s$ increase to $t$ gives that
\begin{equation*} F'(t)\ge\inf_{[x_{i-1},x_{i}]}(f). \end{equation*}
Since $t\in[x_{i-1},x_{i}]$ was arbitrary and $i=1,2,\ldots,n$ was arbitrary we have
\begin{equation*} \inf_{[x_{i-1},x_{i}]}(F')\ge\inf_{[x_{i-1},x_{i}]}(f) \end{equation*}
for all $i=1,2,\ldots,n$. Note that a similar proof can be used to establish this inequality at $t=x_{i-1}$. A similar proof also shows that
\begin{equation*} \sup_{[x_{i-1},x_{i}]}(F')\le\sup_{[x_{i-1},x_{i}]}(f) \end{equation*}
for all $i=1,2,\ldots,n$. We conclude that
\begin{equation*} L(f,P)\le{}L(F',P)\le{}U(F',P)\le{}U(f,P) \end{equation*}
where $L(g,P)$ and $U(g,P)$ denote the lower and upper Darboux sums of $g$ over $P$. Refining the partition and using that $f$ is integrable shows that $F'$ is integrable over $I$.