Properties of the Riemann Integral
Let f be a monotone function on [a,b]. Then f is Riemann integrable.
Direct Proof
Suppose f is increasing on [a,b]. Since f(a) \leq f(x) \leq f(b) for all x \in [a,b], f is bounded on [a,b]. Now given \varepsilon > 0 there exists k > 0 such that
k[f(b)-f(a)] < \varepsilon
By the Arichimeadian property. Let P = \{x_0,\ldots,x_n\} be a partition of [a,b] such that \Delta x_i \leq k for all i (Recall \Delta x_i = x_i - x_{i-1}). Since f is increasing, we have
m_i = f(x_{i-1}) \qquad \text{and} \qquad M_i = f(x_i)
for i \in \{1,\ldots,n\}. Then
\begin{align*} U(f,P) - L(f,P) &= \sum_{i=1}^n[f(x_i)-f(x_{i-1})](\Delta x_i)\\ &\leq k\sum_{i=1}^n[f(x_i)-f(x_{i-1})] = k[f(b)-f(a)] < \varepsilon \end{align*}
Since for all \varepsilon > 0, U(f,P) - L(f,P) < \varepsilon, f is Riemann integrable by Theorem 4.1.4. The proof is similar for decreasing monotone functions.
Let f be a continuous function on [a,b]. Then f is integrable on [a,b].
Direct Proof
Since f is continuous on [a,b], a compact set, it is uniformly continuous on [a,b]. Thus for any \varepsilon > 0, there exists \delta > 0 such that
|f(x)-f(y)| < \frac{\varepsilon}{b-a}
When x,y \in [a,b] and |x-y| < \delta. Let P = \{x_0,\ldots,x_n\} be a partition of [a,b] such that \Delta x_i < \delta for all i \in \{i,\ldots,n\}. Since f assumes its minimum and maximum on [a,b], there exists points s_i,t_i \in [a,b] such that m_i = f(s_i), and M_i = f(t_i). Since x_i - x_{i-1} < \delta, we have that |s_i - t_i| < \delta, and
0 \leq M_i - m_i = f(t_i) - f(s_i) < \frac{\varepsilon}{b-a}
for all i. It follows that
U(f,P) - L(f,P) = \sum_{i=1}^n(M_i - m_i)(\Delta x_i) < \frac{\varepsilon}{b-a}\sum_{i=1}^n\Delta x_i = \varepsilon
Thus by Theorem 4.1.4, f is Riemann integrable on [a,b].
We have now proved that two large classes of functions, monotone functions and continuous functions are Riemann integrable. There of course exists other functions that are neither continuous or monotone, but are still Riemann integrable.
Let f be the Thomae’s function defined on the interval [0,1] as
f(x) = \begin{cases} \frac{1}{n} & \text{ if $x = \frac{m}{n}$ in rational in lowest terms,}\\ 0, & \text{ if $x$ is irrational.} \end{cases}
Let \varepsilon > 0, choose N such that N > 2/\varepsilon. Let Y_N be the set of all rational numbers in [0,1] that have a denominator less than N when expressed in lowest terms, for example
Y_5 = \left\{0,\frac{1}{4},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{3}{4},1\right\}.
Let n > 4m/\varepsilon be an integer, and let P = \{x_0,\ldots,x_n\} such that \Delta x = 1/n for all n.
We want to split our indices into two disjoint subsets based on whether or not the indice contains a point in Y_N. Let
A = \{i \colon [x_{i-1},x_i] \cap Y = \varnothing\} \text{ and } B = \{i \colon [x_{i-1},x_i] \cap Y \not = \varnothing\}.
If i \in A, then every rational in [x_{i-1},x_i] has a denominator greater or equal to N. This means M_i \leq 1/N < \varepsilon/2. It follows that
\sum_{i \in A} M_i \Delta x_i \leq \frac{\varepsilon}{2}\sum_{i \in A} \Delta x_i < \frac{\varepsilon}{2}.
Let f and g be Riemann integerable functions on [a,b] and let k \in \mathbb{R}. Then
(a) kf is Riemann integrable and
\int^b_a kf = k \int^b_a f
(b) f + g is Riemann integrable and
\int^b_a (f+g) = \int^b_a f + \int^b_a g
Direct Proof
(a) If k = 0, then the result is trivial. Suppose k > 0 and let P = \{x_0,\ldots,x_n\} be a partition of [a,b]. For all i we have
M_i(kf) = kM_i(f).
Thus we have U(kf,P) = kU(f,P), and U(kf) = kU(f). Similarly we have L(kf) = kL(f). Since U(f) = L(f),
L(kf) = k L(f) = k U(f) = U(kf).
Thus kf is Riemann integrable and furthermore
\int^b_a kf = U(kf) = kU(f) = k\int^b_a f.
The case where k < 0 is similar.
(b) We know
\sup [(f+g)(D)] \leq \sup f(D) + \sup g(D).
So for any partition P, we have M_i(f+g) \leq M_i(f) + M_i(g) for each i, so
U(f + g,P) \leq U(f,P) + U(g,P).
Similarly
L(f + g, P) \geq L(f,P) + L(g,P).
We want to show U(f+g,P) - L(f+g,P) < \varepsilon. Given any \varepsilon > 0, there exists partitions P_1,P_2 of [a,b] such that
U(f,P_1) < L(f,P_1) + \frac{\varepsilon}{2} \quad \text{and} \quad U(g,P_2) < L(g,P_2) - \frac{\varepsilon}{2}.
Let P = P_1 \cup P_2. Then by Theorem 4.1.2 we have
U(f,P) < L(f,P) + \frac{\varepsilon}{2} \quad \text{and} \quad U(g,P) < L(g,P) - \frac{\varepsilon}{2}.
Combining inequalities we obtain
\begin{align*} U(f+g,P) &\leq U(f,P) + U(g,P)\\ &< L(f,P) + L(g,P) + \varepsilon \leq L(f+g,P) + \varepsilon. \end{align*}
Thus by Theorem 4.1.4 we have that f+g in Riemann integrable. Furthermore since
\begin{align*} \int^b_a(f+g) &= U(f + g) \leq U(f+g,P)\\ &< L(f,P) + L(g,P) + \varepsilon\\ &\leq L(f) + L(g) + \varepsilon = \int^b_a f + \int^b_a g + \varepsilon \end{align*}
and
\begin{align*} \int^b_a(f+g) &= L(f + g) \geq L(f+g,P)\\ &> U(f,P) + U(g,P) - \varepsilon\\ &\leq U(f) + U(g) - \varepsilon = \int^b_a f + \int^b_a g - \varepsilon \end{align*}
we must have
\int^b_a (f+g) = \int^b_a f + \int^b_a g.
If f is Riemann integrable on both [a,c] and [c,b], then f is Riemann integrable on [a,b]. Furthermore
\int^b_a f = \int^c_a f + \int^b_c f.
Direct Proof
Given \varepsilon > 0, there exists partitions P_1 of [a,c] and P_2 of [c,b] such that
U(f,P_1) - L(f,P_1) < \frac{\varepsilon}{2} \quad \text{and} \quad U(f,P_2) - L(f,P_2) < \frac{\varepsilon}{2}.
Let P = P_1 \cup P_2. Then P is a partition of [a,b] and we have
\begin{align*} U(f,P) - L(f,P) &= U(f,P_1) + U(f,P_2) - L(f,P_1) - L(f,P_2)\\ &= [U(f,P_1)-L(f,P_1)] + [U(f,P_2) - L(f,P_2)]\\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{align*}
Thus f is Riemann integrable on [a,b] by Theorem 4.1.4. Furthermore,
\begin{align*} \int^b_a f \leq U(f,P) &= U(f,P_1)+U(f,P_2)\\ &< L(f,P_1) + L(f,P_2) + \varepsilon \leq \int^c_a f + \int^b_c f + \varepsilon, \end{align*}
and
\begin{align*} \int^b_a f \geq L(f,P) &= L(f,P_1)+L(f,P_2)\\ &> U(f,P_1) + U(f,P_2) - \varepsilon \geq \int^c_a f + \int^b_c f - \varepsilon, \end{align*}
Thus
\int^b_a f = \int^c_a f + \int^b_c f.
For any functions f,g which are Riemann integerable on [a,b], such that f(x) \leq g(x) for all x \in [a,b], then
\int^b_a f \leq \int^b_a g
Direct Proof
Later.
Let f be Riemann integerable on [a,b] and g be continuous on [c,d] where f([a,b]) \subseteq [c,d]. Then g \circ f is Riemann integrable on [a,b].
Direct Proof
Given \varepsilon > 0, let K = \sup\{|g(t)| \colon t \in [c,d]\} and choose \varepsilon^\prime > 0 such that \varepsilon^\prime(b-a+2K) < \varepsilon. Since g is continuous on [c,d], it is uniformly continuous on [c,d]. Thus there exists some \delta > 0 such that \delta < \varepsilon^\prime and such that |g(s) - g(t)| < \varepsilon^\prime whenever |s-t| < \delta and s,t \in [c,d]. Since f is integrable on [a,b], there exists some partition P = \{x_0,\ldots,x_n\} of [a,b] such that
U(f,P) - L(f,P) < \delta^2.
We claim that for this partition P we also have
U(g \circ f,P) - L(g \circ f,P) = \sum_{i=1}^n[M_i(g\circ f) - m_i(g \circ f)]\Delta x < \varepsilon.
To show this we seperate our partition P into two disjoint sets. Let
A = \{i \colon M_i(f) - m_i(f) < \delta\} \text{ and } B = \{i\colon M_i(f) - m(i) \geq \delta\}.
Then if i \in A, and x,y \in [x_{i-1},x_i], we have
|f(x) - f(y) | \leq M_i(f) - m_i(f) < \delta
so that |(g \circ f)(x) - (g \circ f)(y) | < \varepsilon^\prime and therefore M_i(g \circ f) - m_i(g \circ f) \leq \varepsilon^\prime. It follows that
\sum_{i \in A}\left[ M_i(g\circ f) - m_i(g \circ f) \right]\Delta x_i \leq \varepsilon^\prime\sum_{i\in A} \Delta x_i \leq \varepsilon^\prime(b-a)
Otherwise, if i \in B, then [M_i(f) - m_i(f)]/\delta \geq 1 so
\begin{align*} \sum_{i \in B} \Delta x_i &\leq \frac{1}{\delta}\sum_{i \in B}[M_i(f) - m_i(f)]\Delta x_i\\ &\leq \frac{1}{\delta}[U(f,P) - L(f,P)] < \delta < \varepsilon^\prime. \end{align*}
Thus since M_i(g \circ f) - m_i(g \circ f) \leq 2K for all i we have
\sum_{i \in B}[M_i(g \circ f) - m_i(g \circ f)]\Delta x \leq 2K\sum_{i \in B}\Delta x < 2K\varepsilon^\prime
Combining our indices we have
\begin{align*} U(g \circ f, P) - L(g \circ f,P) &= \sum_{i \in A}\left[ M_i(g\circ f) - m_i(g \circ f) \right]\Delta x_i\\ &+ \sum_{i \in B}[M_i(g \circ f) - m_i(g \circ f)]\Delta x\\ &\leq \varepsilon^\prime(b-a) + 2K\varepsilon^\prime = \varepsilon^\prime(b-a+2K) < \varepsilon \end{align*}
Hence g \circ f is Riemann integrable on [a,b].
If f is Riemann integrable on both [a,b] and |f| is Riemann integrable on [a,b], then
\left|\int^b_a f \right| \leq \int^b_a |f|
Direct Proof
Since f is Riemann integerable, the function is bounded on [a,b]. Let B \geq |f(x)| for all x \in [a,b]. Define g \colon [-B,B] \to \mathbb{R} such that it maps t \to |t|. Since g is continuous and g \circ f = |f|, we have that |f| is integerable.
Since we have that -|f(x)| \leq f(x) \leq |f(x)| for all x, by Lemma 4.2.5, we have that
- \int^b_a |f| \leq \int^b_a f \leq \int^b_a |f|
which we can then rewrite using the definition of absolute value to obtain
\left| \int^b_a f \right| \leq \int^b_a |f|