The problem: Suppose $a < b$. I want to show that the function $f: [a,b]\to\mathbb{R},\ f(x) = -2x$ is integrable on the interval $[a,b]$, using the fact that a bounded function is integrable, if and only if its lower and upper Riemann-integrals on the interval $[a,b]$ are the same: \begin{align*} \underline{\int_a^b} f &= \sup\left\{ s(P_1) \mid s(P_1) = \sum_{i=1 }^{n} m_i (x_i - x_{i-1})\right\}\\ &= \inf\left\{ S(P_2) \mid S(P_2) = \sum_{j=1 }^{k} M_j (x_j - x_{j-1})\right\}\\ &= \overline{\int_a^b} f, \end{align*}
where $m_i$ and $M_j$ are the infimum and supremum of given subintervals respectively. Obviously $f$ is bounded, so it makes sense to use the above theorem. Based on the definitions of lower- and upper Riemann sums $s(P)$ and $S(P)$, I get that I should choose the partitions $P_1 = \{x_i\}_{i=0}^{n}$ and $P_2 = \{x_j\}_{j=0}^{k}$ so that the sums are equal, but I'm having trouble coming up with appropriate ones. How should I go about this?
I assume that $m_i$ and $M_j$ are supremums of $f$ on a given subinterval? If so, you are talking about Darboux sums, not Riemann sums.
Hint: