$\newcommand{\ds}{\displaystyle}$ $\newcommand{\Rr}{\mathbb{R}}$ $\newcommand{\Qq}{\mathbb{Q}}$ $\newcommand{\lint}{\underline{\int}}$ $\newcommand{\uint}{\overline{\int}}$

Properties of Riemann Integrals¶

Theorem 1.

  1. If $f,g \in R[a,b]$ then $f+g \in R[a,b]$, also, $cf \in R[a,b]$ for every constant $c$. Moreover, $$ \int_a^b f+g = \int_a^b f + \int_a^b g\ \text{and}\ \int_a^b cf = c \int_a^b f. $$

  2. If $f(x) \le g(x)$ on $[a,b]$, then, $\ds \int_a^b f \le \int_a^b g.$

  3. If $f \in R[a,b]$ and if $a < c < b$, then the restrictions of $f$ on $[a,c]$ and on $[c,b]$ are both integrable. Moreover, $$ \int_a^b f = \int_a^c f + \int_c^b f. $$

  4. If $f \in R[a,b]$ and $|f(x)| \le M$ on $[a,b]$, then $$ \left| \int_a^b f\right| \le M |b-a|. $$

The following result is taken from Rudin's Principles of Mathematical Analysis. It is stated and proved there for the more general Riemann-Stieltjes integral.

Theorem 2. Suppose $f$ is integrable on $[a,b]$, $\phi$ continuous on $[m,M]$ and $f([a,b]) \subseteq [m,M]$. Then the composition $h(x) = \phi \circ f$ is integrable on $[a,b]$.

Proof.

Choose $\varepsilon > 0$. Since $\phi$ is continuous on a closed bounded interval, the continuity is uniform. So, there exists $\delta> 0$ such that $|\phi(s)-\phi(t)|< \varepsilon$ whenever $s,t \in [m,M]$ and $|s-t|< \delta$. By choosing a smaller $\delta>0$, if necessary, we can assume $\delta < \varepsilon$.

Since $f$ is integrable on $[a,b]$, there is a partition $P =\{x_0 < x_1 < \cdots < x_n\}$ of $[a,b]$ such that

$$ 0 \le U(P,f)-L(P,f) = \sum_{i=1}^n (M_i-m_i)\Delta x_i < \delta^2. \tag{*} $$ where $M_i, m_i$ have the same meaning as in the definition for Riemann integral. Let $M_i^*, m_i^*$ be the analogous numbers for $h$.

Divide the indices $1, \ldots, n$ into two classes:

$$A = \{i \colon M_i - m_i < \delta \}, B = \{i \colon M_i - m_i \ge \delta\}.$$

For $i \in A$, and any $x,x' \in I_i$, $|f(x)-f(x')| \le M_i-m_i < \delta$. So,

$$|h(x)-h(x')|=|\phi\circ f(x)-\phi\circ f(x')| < \varepsilon.$$

Hence,

$$ M^*_i - m^*_i = \sup\{ |h(x)-h(x')| \colon x,x' \in I_i\} \le \varepsilon. $$

Therefore, $$ \sum_{i \in A} (M_i^* -m_i^*) \Delta x_i \le \varepsilon\sum_{i \in A} \Delta x_i \le \varepsilon(b-a). $$

For $i \in B$, $M_i^*-m_i^* \le 2K$, where $K = \sup\{|\phi(t)| \colon m \le t \le M\}$.

Therefore, it follows from the definition of $B$ and ($*$) that

\begin{align*} \delta \sum_{i \in B} \Delta x_i &\le \sum_{i \in B} (M_i -m_i)\Delta x_i \\ &\le \sum_{i=1}^n (M_i-m_i)\Delta x_i < \delta^2. \end{align*}

Thus, $\sum_{i \in B} \Delta x_i < \delta$. And so,

\begin{align*} U(P,h) - L(P,h) &= \sum_{i=1}^n (M^*_i - m^*_i)\Delta_i \\ &= \sum_{i \in A} (M^*_i - m^*_i)\Delta_i + \sum_{i \in B} (M^*_i - m^*_i)\Delta_i \\ &\le \varepsilon(b-a) + 2K\delta < \varepsilon(b-a+2K). \end{align*}

Since $b-a+2K > 0$ is fixed, $U(P,h) - L(P,h)$ can be arbitrary small for a suitably chosen partition $P$.

Hence, $h$ is integrable.

Proposition 3. If $f,g$ are integrable on $[a,b]$, then so is $fg$.

Proof. By Theorem 1. $f+g$ is also integrable on $[a,b]$. Since $x \mapsto x^2$ is continuous, it follows from Theorem 2 then $(f+g)^2$, $f^2$ and $g^2$ are all integrable on $[a,b]$. Finally, since

$$fg = \frac{1}{2}((f+g)^2 -f^2-g^2),$$

it follows from Theorem 1 that $fg$ is also integrable.

Proposition 4. Suppose $f \in R[a,b]$, then so is $|f|$. Moreover, $$ \left| \int_a^b f \right| \le \int_a^b |f|. $$

Proof. The integrability of $|f|$ follows immediately from the previous theorem with $\phi(x) = |x|$. Since both $f$ and $-f$ are bounded above by $|f|$. By 2. in Theorem 1.

$$ \int_a^b -f, \int_a^b f \le \int_a^b |f|$$

But $\int_a^b -f = -\int_a^b f$ (by 1. of Theorem 1, take $c=-1$), so $\ds \left|\int_a^b f\right| \le \int_a^b |f|$.

Remark. The converse of this proposition is not true as it can be exemplified by the following function (a modification of the Dirichlet function)

$$ f(x) = \begin{cases} \phantom{-}1 &x \in \Qq \cap [0,1] \\ -1 & x \in [0,1] \setminus \Qq. \end{cases} $$

Then it is easy to check that $\ds \overline{\int_0^1} f = 1$ and $\ds \underline{\int_0^1} f = -1$. Hence $f$ is not integrable. However, $|f|$ is simply the function on $[0,1]$ that takes constant value $1$ and so $|f|$ is integrable.

The Class of Riemann Integrable Functions¶

Notation For $n \ge 1$, denoted by $P_n([a,b])$ or just $P_n$ if $[a,b]$ is understood, the uniform $n$-partition of $[a,b]$, i.e. $P_n = \{a =x_0 < x_1 \ldots < x_n=b\}$ where $\Delta x_i = (b-a)/n$ for each $1 \le i \le n$)

Theorem 5. A continuous function on a closed bounded interval is integrable.

Proof. Take an $\varepsilon >0$. By uniform continuity (a continuous on closed bounded interval implies uniform continuity), there exists $\delta >0$ such that whenever $x,x' \in [a,b]$ with $|x-x'| < \delta$,

$$ |f(x)-f(x')| < \frac{\varepsilon}{b-a} \tag{*}. $$

Take $n$ sufficiently large so that $(b-a)/n < \delta$. We will show that $U(P_n,f)-L(P_n,f) < \varepsilon$.

For each $1 \le i \le n$, by the EVT, $M_i=f(x_i^*)$ and $m_i=f(x_i^{**})$ for some $x_i^*, x_i^{**} \in I_i$. So, $|x_i^*-x_i^{**}| \le (b-a)/n < \delta$, and hence

$$ M_i -m_i = f(x_i^*)-f(x_i^{**}) = |f(x_i^*)-f(x_i^{**})| < \frac{\varepsilon}{b-a}. $$

Thus, $$ U(P_n,f) - L(P_n,f) = \sum (M_i -m_i)\frac{b-a}{n} < \frac{\varepsilon}{b-a} (b-a) = \varepsilon. $$

This shows that $f$ is integrable on $[a,b]$.

Theorem 6. A monotonic function on a closed bounded interval is integrable.

Proof. Suppose $f$ increasing on $[a,b]$. Then, for $P_n$ ($n \ge 1$), we have $M_i = f(x_i)$ and $m_i = f(x_{i-1})$ and so $$ U(P_n,f)-L(P_n,f) = \sum (f(x_i)-f(x_{i-1}))\Delta x_i = \frac{b-a}{n}(f(b)-f(a)). $$ The last quantity tends to $0$ and $n \to \infty$. Thus, $f \in R[a,b]$.

Example/Exercise. Thomae (raindrop) function

The function $h \colon [0,1] \to \Rr$ defined by

$$ h(x) \begin{cases} \dfrac{1}{n} & \text{if}\ x = \dfrac{m}{n}\ \text{where} \gcd(m,n)=1 \\ \\ 0 & \text{if}\ x \in [0,1] \setminus \mathbb{Q} \end{cases} $$ is a integrable on $[0,1]$ but neither continuous nor monotone on any subinterval of $[0,1]$. $\int_0^1 h = 0$.

Proof.

Since both rational numbers and irrational numbers are dense, $h$ is not monotonic on any subinterval.

Let $q$ be a rational number in $[0,1]$. Take a sequence $(\alpha_n)$ of irrational numbers in $[0,1]$ that converging to $q$ (irrational numbers are dense). If $h$ were continuous at $q$ then we would have $h(\alpha_n) = 0 \to h(q) > 0$, contradiction. So, $h$ must be discontinuous at the rational numbers and since the rational numbers are dense in the reals, $h$ is not continuous on any subinterval of $[0,1]$.

The lower integral of $h$ is $0$ since the irrationals are dense. To see that $h$ is integrable, it suffices to show that its upper sums can be arbitraily small.

For $N > 1$, let

$$H_N = \bigcup_{1 \le k \le N} h^{-1}(1/k) = \left\{x \in [0,1] \colon h(x) \ge 1/N \right\}$$

Note that $|h^{-1}(1/k)| \le k$ and so the size of $H_N$ is $O(N^2)$. Now take a partition $P$ of $[0,1]$ so that the points in $H_N$ are contained in distinct subintervals whose widths are less than $1/N^3$. On those subintervals that do not contain points of $H_N$, the value of $h$ is at most $1/N$. Thus, the upper sum $U(P,f)$ is $O(N^2)/N^3 + O(1/N) = O(1/N)$. Therefore, $h$ is integrable and $\ds \int_0^1 h = 0$.