(a) $\overline {\int_0^1} x dx = \left(\frac{1}{n}\right)\left(0 + \frac{1}{n} + \frac{2}{n} + \dots + \frac{n - 1}{n}\right) = \frac{n(n-1)/2}{n^2} = \frac{n - 1}{2n}$, so the upper integral $\leq \frac{1}{2}$ as $n \to \infty$
$\underline{\int_0^1} x dx = \left(\frac{1}{n}\right)\left( \frac{1}{n} + \dots + 1 \right) = \frac{n(n + 1)/2}{n^2} = \frac{1}{2}$, so the lower integral $\geq \frac{1}{2}$ as $n \to \infty$
Since the upper integral is always greater than the lower integral, they must both be $\frac{1}{2}$.
So the lower and upper sums are equal, so the $x$ is integrable on $[0, 1]$ and its integral is $\frac{1}{2}$.
Since $g \leq f \leq h$,