= Solution
Fix $x\in(0,1)$. For any sufficiently small nonzero $h$, the additivity of the <Riemann integral> gives
$$
\frac{F(x+h)-F(x)}h-f(x)
=\frac1h\int_x^{x+h}[f(t)-f(x)]\,dt.
$$
Taking absolute values, for either sign of $h$,
$$
\left|\frac{F(x+h)-F(x)}h-f(x)\right|
\leq\sup_{|t-x|\leq|h|}|f(t)-f(x)|.
$$
Because $f$ is <continuous> at $x$, the right side tends to zero. Thus the <fundamental theorem of calculus> here follows directly from the difference quotient:
$$
\boxed{F'(x)=f(x)\quad(0<x<1).}
$$
Only <continuity> at the point under consideration was used.
Without <continuity>, the assertion is false. Take the <Riemann integrable> step function $f(t)=0$ for $t<1/2$ and $f(t)=1$ for $t\geq1/2$. Its primitive is $F(x)=0$ for $x\leq1/2$ and $F(x)=x-1/2$ for $x\geq1/2$. The left and right difference quotients at $1/2$ are respectively zero and one, so \b[the primitive need not be differentiable at every interior point]. A function with this single jump is <Riemann integrable>; changing its value at the jump does not change the <integral> or the counterexample.
Back to article page
