# Friday, February 15, 2013 Riemann Criterion, and some results about integrability

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Riemann Criterion for Integrability

Let $I = [a,b]$ and $f : I \to \mathbb{R}$ be a bounded function on $I$. $f$ is Riemann integrable on $I \iff \epsilon > 0$ given there exists a partition $P_\epsilon$ of $I$ such that $U_f(P_\epsilon) - L_f(P_\epsilon) < \epsilon$

Proof: ($\Rightarrow$) if $f$ is Riemann integrable, then $L(f) = U(f)$. $U(f) = \inf\{U_f(P):P \text{ is a partition of } I\}$. If $\epsilon > 0$ is given, then $U(f) + \epsilon / 2$ is no longer a lower bound. Thus there exists a partition $P_1$ of $I$ such that $U_f(P_1) < U(f) + \epsilon / 2$. Similarly, by the properties of the supremum, there exists a partition $P_2$ of $I$ such that $L(f) - \epsilon / 2 < L_f(P_2)$. Let $P_\epsilon = P_1 \cup P_2$, a refinement of $P_1$ and $P_2$. Thus $L(f) - \epsilon / 2 < L_f(P_2) \leq L_f(P_\epsilon)$ and $U_f(P_\epsilon) \leq U_f(P_1) < U(f) + \epsilon / 2$. Thus $U_f(P_\epsilon) - L_f(P_\epsilon) < U(f) + \epsilon / 2 - (L(f) - \epsilon / 2) = \epsilon$.

($\Leftarrow$) Let $\epsilon > 0$ be given, there exists $P_\epsilon$ a partition of $I$ such that $U_f(P_\epsilon) - L_f(P_\epsilon) < \epsilon$. But as you know, $L_f(P_\epsilon) \leq L(f) \leq U(f) \leq U_f(P_\epsilon)$ so $U(f) - L(f) \leq U_f(P_\epsilon) - L_f(P_\epsilon) < \epsilon$, i.e. $0 \leq U(f) - L(f) < \epsilon$ for all $\epsilon > 0$ thus $U(f) - L(f) = 0$. Thus $U(f) = L(f)$ and $f$ is Riemann integrable. $\blacksquare$

Corollary: Let $I = [a,b]$ and $f : I \to \mathbb{R}$ be a bounded function on $I$. If $\{P_n\}$ is a sequence of partitions of $I$ such that $\displaystyle \lim_{n \to \infty} U_f(P_n) - L_f(P_n) = 0$ then $f$ is Riemann integrable on $I$ and $\displaystyle\int_a^b f = \lim_{n \to \infty} U_f(P_n) = \lim_{n \to \infty} L_f(P_n)$.

Proof: If $\epsilon > 0$ is given there exists some $N \in \mathbb{N}$ such that if $n \geq N$, $U_f(P_n) - L_f(P_n) < \epsilon$ (as $\lim U_f(P_n) - L_f(P_n) = 0$). Thus by Riemann Criterion, $f$ is integrable on $I$. To show that $\displaystyle\int_a^b f = \lim_{n \to \infty} U_f(P_n)$ we note that $L_f(P_n) \leq L(f) \leq U(f) \leq U_f(P_n)$ thus $0 \leq U_f(P_n) - U(f) \leq U_f(P_n) - L_f(P_n) < \epsilon$ for $n \geq N$ thus $\lim U_f(P_n) = U(f) = \int_a^b f$. The proof follows similarly for $\lim L_f(P_n) = L(f) = \int_a^b f$.

Theorem: Let $I = [a,b]$ and $f : I \to \mathbb{R}$ be a monotone function, then $f$ is Riemann integrable on $I$.

Proof: Let $f$ be monotone increasing and let $P_n = \{ x_0, x_1, \ldots, x_n \}$ be a regular partition of $I$ into $n$ equal parts of length $\frac{b - a}{n}$. Since $f$ is monotone increasing,

$$\sup \{ f(x) : x \in [x_{k-1}, x_k] \} = f(x_k) \text{ and } \inf \{ f(x) : x \in [x_{k-1}, x_k] \} = f(x_{k-1})$$

so

\begin{align} U_f(P_n) &= \sum_{k=1}^n f(x_k)(x_k - x_{k-1}) \\ &= \sum_{k=1}^n f(x_k)\frac{b-a}{n} \\ &= \left(\frac{b-a}{n}\right)(f(x_1) + f(x_2) + \ldots + f(x_n)) \end{align}

and similarly $\displaystyle L_f(P_n) = \left(\frac{b-a}{n}\right)(f(x_0) + f(x_2) + \ldots + f(x_{n-1}))$ so $U_f(P_n) - L_f(P_n) = \left(\frac{b-a}{n}\right)(f(x_n) - f(x_0)) = \left(\frac{b-a}{n}\right)(f(b) - f(a))$. For any given $\epsilon > 0$, for $\displaystyle n > \frac{(b-a)(f(b)-f(a))}{\epsilon}$ it follows that

$$\frac{(b-a)(f(b)-f(a))}{n} < \epsilon$$

i.e. $U_f(P_n) - L_f(P_n) < \epsilon$ so $f$ is Riemann integrable on $I$. The proof for $f$ monotone decreasing follows similarly.

Theorem: Let $I = [a,b]$ and $f : I \to \mathbb{R}$ continuous on $I$. Then $f$ is Riemann integrable on $I$.

Proof: As $f$ is continuous on a closed and bounded interval $I$ then $f$ is uniformly continuous on $I$. Thus for a given $\epsilon > 0$ there exists some $\delta > 0$ such that if $|x - y| < \delta$ then $|f(x) - f(y)| < \epsilon / (b - a)$. Let $P_n = \{ x_0, x_1, \ldots, x_n \}$ be a regular partition of $I$ into subintervals of length $(b-a) / n$. Then for $n > \frac{b-a}{\delta}$ (i.e., making sure that $\frac{b-a}{n} < \delta$) $f$ is continuous on $[x_{k-1}, x_k]$ and there exist $u_k, v_k \in [x_{k-1}, x_k]$ such that $M_k = \sup\{f(x) : x \in [x_{k-1}, x_k]\} = f(v_k)$ and $m_k = \inf\{f(x) : x \in [x_{k-1}, x_k]\} = f(u_k)$. Then $|u_k - v_k| < \delta$ thus

$$|f(u_k) - f(v_k)| = M_k - m_k < \frac{\epsilon}{b-a}$$
thus

\begin{align} U_f(P_n) - L_f(P_n) &= \sum_{k=1}^n M_k(x_k - x_{k-1}) - \sum_{k=1}^n m_k(x_k - x_{k-1}) \\ &= \sum_{k=1}^n (M_k-m_k)(x_k - x_{k-1}) \\ &< \sum_{k=1}^n \frac{\epsilon}{b-a}(x_k - x_{k-1}) \\ &= \frac{\epsilon}{b-a} \sum_{k=1}^n x_k - x_{k-1} \\ &= \frac{\epsilon}{b-a} (x_n - x_0) = \frac{\epsilon}{b-a}(b - a) = \epsilon \end{align}