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 $I=\int^b_af$.

Answer 2

$\newcommand{\d}{\,\mathrm{d}}\newcommand{\P}{\mathscr{P}}$A bounded function $f:[a,b]\to\Bbb R$ is Riemann integrable if and only if (various equivalent definitions, Spivak's is equivalent to this one) there exists $J\in\Bbb R$ such that for all $\epsilon>0$ there exists $\delta>0$ such that for all tagged partitions $\P$ of $[a,b]$ with mesh less than $\delta$, we have: $$\left|\sum_{(x,y,t)\in\mathscr{P}}(y-x)f(t)-J\right|<\epsilon$$

In which case, we define: $$\int_a^bf:=J$$

In particular, we know $x\mapsto e^x$ is Riemann integrable on $[0,1]$. Consider the tagged partitions $\P_n$ which have elements $((k-1)/n,k/n,k/n)$ for $1\le k\le n$. I know that for all $\epsilon>0$ there is $\delta>0$ such that, if the mesh $\|\P_n\|=1/n$ is less then $\delta$, then: $$\tag{$\ast$}\int_0^1e^x\d x-\epsilon<\frac{1}{n}\sum_{k=1}^ne^{k/n}=\sum_{(x,y,t)\in\P_n}(y-x)e^t<\int_0^1e^x\d x+\epsilon$$

But of course, if $N:=\lceil\delta^{-1}\rceil+1$ then for all $n\ge N$ we have $(\ast)$ since $1/n<\delta$. By definition of the limit of a sequence indexed in $\Bbb N$, it follows that: $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^ne^{k/n}=\int_a^b e^x\d x$$

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How can we relate that to the Darboux concept of $U$ and $L$? A full proof is in the Wikipedia page. But I'll sketch the idea. As you observe, we have: $$\frac{1}{n}\sum_{k=1}^ne^{k/n}=U(f;\P_n)$$With $f:x\mapsto e^x$ defined on $[0,1]$. I know: $$\inf_{\P}U(f;\P)=\int_0^1e^x\d x$$

But notice that if $\P$ is a refinement of $\P'$, then $U(f;\P)\le U(f;\P')$. Moreover, the $\inf$ of a set of real numbers can be attained as a sequence $a_1\ge a_2\ge\cdots$ tending to the $\inf$. You can imagine taking, for large $n$, those special $\P=a_n$ which have $U(f;\P)$ being close to $\int_0^1 e^x\d x$. But for very large $n$ it is true that $U(f;\P)\approx U(f;\P_n)$ since the $0,1/n,2/n,\cdots$ partition has sufficiently fine divisions. So you can obtain $U(f;\P_n)\to\int_0^1e^x\d x$.

Answer 3

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

The answer is "yes" by definition. Your "definition" left out the part where you actually define something: if the the infimum and supremum are equal, then their common value is the value of the integral. From the implicit "only if" nature of a definition, we can then say that if the integral exists, it must be true that the infimum and supremum are equal. Therefore you can write

$$\text{if } \int_a^b f \text{ is defined then } \lim_{n\to\infty} U(f,P_n)=\lim_{n\to\infty} L(f,P_n)=\int_a^b f.$$

In this particular solution to this particular exercise, you need to know that $\int_0^1 e^x\,\mathrm{d}x$ exists. You could, of course, go all the way back to the definition of an integral and prove this for this particular case, but part of applying the definition would be solving the exercise. Instead, the published solution simply recognizes $e^x$ as a function that is integrable on $[0,1].$

How does Spivak know that $e^x$ is integrable on $[0,1]$? Spivak is a careful worker, so I expect this particular fact was established earlier, or at least it has been established that $e^x$ is a continuous function of $x,$ which along with bounds of $e^x$ over $[0,1]$ guarantees that the integral exists via a theorem that is proved somewhere earlier in the text.

Answer 4

To address the specific question at the end:

If $f$ is Riemann integrable with integral value $I_0$, then by definition we have

$$I_0 = \underline{\int}_a^b f := \sup_PL(f,P) = \overline{\int_a}^bf := \inf_P U(f,P)$$ By definition of the infimum and supremum, for any $n \in \mathbb{N}$ there exist partitions $P_n'$ and $P_n''$ such that

$$I_0 - \frac{1}{n} < L(f,P_n') \leqslant I_0 \leqslant U(f, P_n'') < I_0 + \frac{1}{n}$$

Taking the common refinement $P_n = P_n' \cup P_n''$, we have

$$L(f,P_n') \leqslant L(f,P_n) \leqslant I_0 \leqslant U(f,P_n) \leqslant U(f,P_n'')$$

Hence,

$$I_0 - \frac{1}{n} \leqslant L(f,P_n) \leqslant I_0 \leqslant U(f,P_n) \leqslant I_0 + \frac{1}{n},$$

and, therefore, there exists a sequence of partitions $(P_n)$ for which $$\lim_{n \to \infty} U(f,P_n) = \lim_{n \to \infty} L(f,P_n) = I_0$$