When does the upper sum and lower sum converge to the definite integral

Question

I am reading Spivak's Calculus and in problem 9 in Chapter 22, we are asked to evaluate the limit $\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n e^{k/n}$. Spivak simply puts this equal to $\int_0^1 e^x\mathrm{d}x=e-1$, since the sum is the upper sum of the integral. But nowhere has he stated a sufficient condition for when the upper sum $U(f,P)$ (or lower sum $L(f,P)$) converges to the definite integral.

His definition of the definite integral $\int_a^b f$ is when $$\mathrm{inf} \{ U(f,P): P \ \text{a partition of} \ [a,b] \}=\mathrm{sup} \{ L(f,P): P \ \text{a partition of} \ [a,b] \}.$$ From this it follows that $L(f,P)\leq \int_a^b f \leq U(f,P)$ for all partitions $P$ of $[a,b]$.

My question then. For a partition $P_n$ of $n$ points, is there an implication (or even an equivalence) $$\text{if} \ldots \text{then} \lim_{n\to\infty} U(f,P_n)=\lim_{n\to\infty} L(f,P_n)=\int_a^b f.$$

Answer 1

I don't remember how the Riemann integral is defined in Spivak's (whether through the use of refinement of partitions or through taking limits other the size of the partition $\|P\|\rightarrow0$). In any event, using refinements or limits of sizes of partitions turns out to be equivalent.
This is a well know result in Riemann integration (See here, page 132 for example)

Proposition A: For any bounded function $f$ on $[a,b]$ \begin{align} \lim_{\|P\|\rightarrow0}U(f,P)&=\inf_PU(f,P)\\ \lim_{\|P\|\rightarrow0}L(f,P)&=\sup_PL(f,P) \end{align}

As a consequence of Prop A, if $f$ is known to be Riemann integrable over $[a,b]$, for any sequence of partitions $P_n=\{x_k=a+\frac{b-a}{n}k, \,0\leq k\leq n\}$, then $$\lim_{n\rightarrow \infty}\frac{b-a}{n}\sum^n_{k=1}f(x_k)=\int^b_af$$.

Your function $f(x)=e^x$, being continuous, is Riemann integrable in $[0,1]$, and the latter applies.

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Proposition A above also implies the following

Proposition B: Let $P_n=\{a+\tfrac{k(b-a)}{n}:0\leq k\leq n\}$. A bounded function $f$ on $[a,b]$ is Riemann integrable iff $$I=\lim_{n\rightarrow\infty}\frac{b-a}{n}\sum^n_{k=1}f(t_{nk})$$ exists and is independent of the tags $\tau_n=\{t_{nk}:1\leq k\leq n\}$ with $a+\frac{(k-1)(b-a)}{n}\leq t_{nk}\leq a+\tfrac{k(b-a)}{n}$. The common limit is given by $I=\int^b_af$.