How to show that the upper and lower Riemann integrals of a function, say $f(x) = -2x$, are equal?

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?

• Is it obvious they should be equal? Can you not just show you can make the upper and lower as close as you like to each other? – snulty Feb 1 '18 at 14:30
• @Did I'm using Trench's book: digitalcommons.trinity.edu/mono/7, where in Theorem 3.2.6 he uses the "="-sign between the upper and lower integrals. – SeSodesa Feb 1 '18 at 14:39
• Yeah, except that the supremum over every lower-integral and the infimum over every upper-integral are equal, not the lower-integral and the upper-integral for a given partition. – Did Feb 1 '18 at 14:58

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.
• You don't need to find partition for which the sums are equal, only similar (i.e. $S(P)-s(P)<\epsilon$).
• You can easily calculate the values $M_j$ and $m_i$ since $f$ is a monotone function.
• Alright, so I could simply use a partition $P$ whose norm $\lVert P \rVert = \frac{b-a} n \to 0$ as $n\to\infty$, making it possible for the difference $S(P) - s(P)$ to approach zero, effectively making it less than $\varepsilon$. – SeSodesa Feb 1 '18 at 14:55
Compute the lower and the upper sum for an equidistant partition into $N\gg1$ parts exactly and verify that their difference can be made arbitrarily small by choosing $N$ large enough. Then use the squeeze theorem on the relation $$\underline{S}_N\leq\underline{\int_a^b} f(x)\>dx\leq\overline{\int_a^b} f(x)\>dx\leq\overline{S}_N\ .$$