Exercise 7.2.3

Question

[Sequential Criterion for Integrability]
\begin{enumerate}[label=(\alph*)] 
\item Prove that a bounded function $f$ is integrable on $[a, b]$ if and only if there exists a sequence of partitions $\left(P_{n}\right)_{n=1}^{\infty}$ satisfying
$$
\lim _{n \rightarrow \infty}\left[U\left(f, P_{n}\right)-L\left(f, P_{n}\right)\right]=0,
$$
and in this case $\int_{a}^{b} f=\lim _{n \rightarrow \infty} U\left(f, P_{n}\right)=\lim _{n \rightarrow \infty} L\left(f, P_{n}\right)$.
\item For each $n$, let $P_{n}$ be the partition of $[0,1]$ into $n$ equal subintervals. Find formulas for $U\left(f, P_{n}\right)$ and $L\left(f, P_{n}\right)$ if $f(x)=x$. The formula $1+2+3+$ $\cdots+n=n(n+1) / 2$ will be useful.
\item Use the sequential criterion for integrability from (a) to show directly that $f(x)=x$ is integrable on $[0,1]$ and compute $\int_{0}^{1} f$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item $(\implies)$ If $f$ is integrable, then we can choose $P_{Un}$ to satisfy $U(f, P_{Un}) - U(f) < 1/n$, $P_{Ln}$ to satisfy $L(f) - L(f, P_{Ln}) < 1/n$, and $P_n$ to be the common refimenemt of $P_{Un}$ and $P_{Ln}$; it's easy to show in this case that $lim_{n \to \infty} U(f, P_n) = U(f) = \int^b_a f = L(f) = \lim_{n \to \infty} L(f, P_n) $.

$(\impliedby)$ Consider $\lim_{n\to\infty} U(f,P_n) - L(f) \leq U(f,P_n) - L(f,P_n)$. By the Squeeze Theorem, $U(f,P_n) - L(f)$  approaches 0, and therefore $\lim_{n\to\infty} U(f,P_n) = L(f)$. A similar argument shows that $\lim_{n\to\infty} L(f,P_n) = U(f)$. Applying the Algebraic Limit Theorem gets us that $L(f) - U(f) = 0$, or that $L(f) = U(f) = \int^b_a f$ by definition, with this being equal to $lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} L(f, P_n) $.

\item Let $[a_{ni}, b_{ni}]$ be the $i$'th partition of $P_n$ (indexing from 0), with $a_{ni} = i/n$ and $b_{ni} = (i+1)/n$. It is easy to see that
    $$U(f,P_n) = \sum^{n-1}_{i=0} \frac{b_{ni}}{n} = \sum^{n-1}_{i=0} \frac{i+1}{n^2} = \frac{1}{2} + \frac{1}{2n}$$
    and
    $$L(f,P_n) = \sum^{n-1}_{i=0} \frac{a_{ni}}{n} = \sum^{n-1}_{i=0} \frac{i}{n^2} = \frac{1}{2} - \frac{1}{2n}$$

\item $U(f,P_n)$ and $L(f,P_n)$ both approach $1/2$ as $n \to \infty$ and therefore $\int_0^1 f = 1/2$.
 \end{enumerate}

Exercise 7.2.3

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Exercise 7.2.3

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