Let $I = \int_a^b f$ and fix $\varepsilon > 0$. From the definition of Riemann integrability, there are partitions $Q_1, Q_2$ of $[a, b]$ such that
$$I-\varepsilon \leqslant \underline{\Sigma}(f, Q_1) \leqslant \overline{\Sigma}(f, Q_2) \leqslant I + \varepsilon.$$
By letting $Q = \{ t_0, t_1, \ldots, t_m \}$ (where $a = t_0 < t_1 < \cdots < t_m = b$) be the common refinement of $Q_1, Q_2$, we get
$$I-\varepsilon \leqslant \underline{\Sigma}(f, Q) \leqslant \overline{\Sigma}(f, Q) \leqslant I + \varepsilon.$$
Since $f$ is bounded, there is $M$ such that $|f(x)| \leqslant M$ for all $x \in [a, b]$. Let $p$ be so large that
(i) $(m-1) \cdot \frac{b-a}{p} \cdot M \leqslant \varepsilon$,
(ii) $\frac{b-a}{p} \leqslant \displaystyle \min_{1 \leqslant i \leqslant m} (t_i - t_{i-1})$.
Now fix $n \geqslant p$. We will show that
$$I - 3\varepsilon \leqslant \underline{\Sigma}(f, P_n) \leqslant \overline{\Sigma}(f, P_n) \leqslant I + 3\varepsilon,$$
which will complete the proof.
Let $R$ be the common refinement of $Q$ and $P_n$. We're going to show that $R$ has just a little greater inferior sum than $P_n$. By (ii) each interval in $P_n$ contains at most one point $t_i$, unless it contains two, in which case they are it's endpoints. So for each $k = 1, 2, \ldots, m-1$: either $t_k \in P_n$, or adding $t_k$ to $P_n$ divides some interval $I_k$ in $P_n$ into two proper subintervals $I_k = I_k^- \cup I_k^+$ (and $I_k \neq I_l$ for $k \neq l$). Let $S$ be the set of those $k$ for which the second case holds. Then it's easy to see that
$$\begin{align*}
\underline{\Sigma}(f, R) - \underline{\Sigma}(f, P_n) & = \sum_{k \in S} \left[ |I_k^-| \cdot \inf_{x \in I_k^-} f(x) + |I_k^+| \cdot \inf_{x \in I_k^+} f(x) - |I_k| \cdot \inf_{x \in I_k} f(x) \right] \\[1ex]
& \leqslant \sum_{k \in S} \left[ |I_k^-| \cdot M + |I_k^+| \cdot M + |I_k| \cdot M \right] = 2M \sum_{k \in S} |I_k|.
\end{align*} $$
But of course there are at most $m-1$ points in $S$ and $|I_k| = \frac{b-a}{n} \leqslant \frac{b-a}{p}$. Hence, by (i)
$$\underline{\Sigma}(f, R) - \underline{\Sigma}(f, P_n) \leqslant 2M \cdot (m-1) \cdot \frac{b-a}{p} \leqslant 2 \varepsilon.$$
Also since $R$ is a refinement of $Q$, $\underline{\Sigma}(f, R) \geqslant \underline{\Sigma}(f, Q)$. So we get
$$\underline{\Sigma}(f, P_n) \geqslant \underline{\Sigma}(f, R) - 2\varepsilon \geqslant \underline{\Sigma}(f, Q) - 2 \varepsilon \geqslant I - 3\varepsilon.$$
We analogously show that $\overline{\Sigma}(f, P_n) \leqslant I + 3 \varepsilon$, which concludes the proof.
Since $\mathbb{Q}$ and $\mathbb{R} \backslash \mathbb{Q}$ are dense. In each Interval there $[a_j,a_{j+1}]$ $\sup\{f(x):x \in [a_j,a_{j+1}]\}= a_{j+1}$ and $\inf\{f(x):x \in [a_j,a_{j+1}]\}= -a_{j+1}$ Therefore
$\overline{\int_{0}^{t}}f dx=t^2/2$,
$\underline{\int_{0}^{t}}fdx=-t^2/2$
Lets assume we want to calculate $\overline{\int_{0}^{t}}f (x) dx$. We take the partition $Z_n=[0,t/n,(2t)/n,...,t]$ note that if $a_j$ denotes the jth term in the partition. $a_j-a_{j+1}=t/n$ and $ t/n\to 0$ for $n\to \infty$ since the supremum of $f$ on each interval is $a_{j+1}=t(j+1)/n$ the integral equals $\lim_{n \to \infty}\sum_{j=1}^n (t/n) \frac{t(j+1)}{n}=t^2/2$
Best Answer
For that function, each superior sum is equal to $1$, and each inferior sum is equal to $0$. Therefore, take any $\varepsilon\in(0,1]$.