 Solution

Put $G(t)={\int }_a^{t}f(x)dx$. At an interior point and for $h\ne 0$ small enough,By continuity at $t$, the modulus is bounded by the supremum of $|f(x)-f(t)|$ between $t$ and $t+h$, which tends to zero. Thus the fundamental theorem of calculus givesAt the interval endpoints the same argument supplies the corresponding one-sided derivatives.

For an integrable discontinuous function, the single-point spike from part (i) has $G\equiv 0$, so $G^{\prime }(c)=0\ne f(c)=1$; this disproves the derivative identity even when the primitive is differentiable. To see that differentiability itself may fail, take the step function $f(x)=0$ for $x<c$ and $f(x)=1$ for $x\ge c$. It is Riemann integrable, but its primitive is $G(t)=\max (0,t-c)$ and has left derivative zero and right derivative one at $c$. Hence the full assertion is false for general Riemann-integrable functions.