Consider the piecewise function $f$ bounded on $[-1,1]$ given by
$$f(x):= \begin{cases} 1 \text{ if } x>0\\ 0 \text{ if } x\leq0\\ \end{cases} $$
Let $P=\left\{-1,-\frac{3}{4},-\frac{1}{2},-\frac{1}{4},0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1\right\}$ be the partition of $[-1,1]$. Then
$$ m_{j}=\inf \left\{f(x):x_{j-1}\leq x\leq x_{j}\right\}= M_{j}=\sup\left\{f(x):x_{j-1}\leq x\leq x_{j}\right\}=\begin{cases} 1\text{ for } x\in(0,1]\\0 \text{ otherwise } \end{cases} $$
Then the lower Darboux sum is given by
$$ U(P,f)=\sum_{j=1}^{n} m_{j}\Delta x_{j}=0+0+0+0+1(x_{5}-x_{4})+1(x_{6}-x_{5})+1(x_{7}-x_{6})+1(x_{8}-x_{7})=1(x_{8}-x_{4})=1 $$
Similarly, the upper Darboux sum is given by
$$ U(P,f)=\sum_{j=1}^{n} M_{j}\Delta x_{j}=0+0+0+0+1(x_{5}-x_{4})+1(x_{6}-x_{5})+1(x_{7}-x_{6})+1(x_{8}-x_{7})=1(x_{8}-x_{4})=1 $$
Since both the upper sum and lower sum are bounded, and since the least upper bound and greatest lower bound of the Darboux sums is 1, it follows that the lower Darboux integral and the upper Darboux integral are identically 1. So $f\in\textit{R}[-1,1]$ and $\int_{-1}^{1} f(x)dx=1$
This is not quite correct since $\inf \{f(x) \,|\, x_4 \leqslant x \leqslant x_5\}= \inf \{f(x) \,|\, 0 \leqslant x \leqslant 1/4\} = 0$, and we can't conclude that the lower Darboux sum equals the upper Darboux sum for this or any other partition.
However, in the same way we can choose a simpler partition $P_\epsilon=(-1, 0, \epsilon, 1)$ where
$$1- \epsilon = L(P_\epsilon,f) \leqslant U(P_\epsilon,f) = 1$$
Thus, $U(P_\epsilon,f) - L(P_\epsilon,f) < \epsilon$ which immediately proves integrability by the Riemann criterion.
Also, we must have
$$1- \epsilon = L(P_\epsilon,f) \leqslant \sup_P L(P,f) \leqslant \inf_P U(P,f) \leqslant U(P_\epsilon,f) \leqslant 1$$
Since this holds for all $\epsilon >0$, it follows that the upper and lower Darboux integrals both equal $1$, and, hence, the Riemann integral equals $1$.