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A left-continuous cumulative function of an atomic measure with positive masses at a dense countable set has dense discontinuities. Nevertheless it is bounded and monotone on a compact interval, hence a Riemann-integrable function. Finite step-function sums approximate it uniformly because the positive masses are summable. Thus dense discontinuities do not imply failure of the Riemann integral.

For a nonnegative Riemann-integrable function and an interval partition,Every factor is nonnegative, so multiplication of $1+t\le e^t$ is legitimate. The final step uses the lower-sum bound on the Riemann integral.

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.

Changing the value of a bounded Riemann-integrable function at one point preserves its Riemann integral. The difference is supported at that point: partitions isolating it in intervals of arbitrarily small total length make both Darboux sums tend to zero. Thus a nonnegative single-point spike can have zero integral without vanishing everywhere. Its indefinite integral is identically zero, so the fundamental theorem of calculus derivative identity can fail at the changed point despite differentiability of the primitive.