Let $a,b\in\mathbb R$, with $a<b$, and let $f\colon[a,b]\longrightarrow\mathbb R$ be a Riemann integrable function. For each $n\in\mathbb N$, let$$P_n=\left\{a,a+\frac1n(b-a),a+\frac2n(b-a),\cdots,a+\frac{n-1}n(b-a),a+\frac nn(b-a)=b\right\}.$$
Prove that$$(\forall\varepsilon>0)(\exists p\in\mathbb{N}):n\geqslant p\implies\overline{\sum}(f,P_n)-\underline{\sum}(f,P_n)<\varepsilon,\tag{1}$$where, of course, $\overline\sum$ and $\underline\sum$ denote the superior sum and the inferior sum respectively.
Notes:
- If $f$ is continuous, this is easy to prove, using uniform continuity.
- It is also easy to prove that, if $f$ is bounded and $(1)$ holds, then $f$ is Riemann integrable.
- One criterion for Riemann integrability states that (assuming that $f$ is bounded) $f$ is Riemann integrable if and only if, for every $\varepsilon>0$, the inequality $\overline{\Sigma}(f,P)-\underline{\Sigma}(f,P)<\varepsilon$ holds for some partition $P$. So, the problem here consists in proving that it is enough to prove it for those partitions with respect to which all intervals have the same length.
Best Answer
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.