Riemann Integral

The Riemann Integral was the first formal definition of an integral, defined by Bernhard Riemann in 1854. Gaston Darboux developed a simpler approach to the Riemann Integral in 1875 (source needed), that I’ll be using in this section.

Partitions

Definition

Let [a,b] be an interval in \mathbb{R}. A partition P of [a,b] is a finite set of points \{x_0,\ldots,x_n\} in [a,b] such that

a = x_0 < x_1 < \cdots < x_n = b.

A partition serves to chop up our interval [a,b] into fintely many subintervals [x_{i-1},x_{i}]. We notate the length of the i^{th} sub-interval in our partition with \Delta x_i = x_i - x_{i-1}.

Definition

If P and Q are two partitions of [a,b] with P \subseteq Q, then Q is called a refinement of P.

Darboux sums

Definition

Let f be a bounded function on [a,b] and P = \{x_0,\ldots,x_n\} be any partition on [a,b]. Let

M_i(f) = \sup\{f(x) \colon x \in [x_{i-1},x_i]\} and m_i(f) = \inf\{f(x) \colon x \in [x_{i-1},x_i]\}

for each i \in \{1,\ldots,n\}. We define U(f,P) = \sum^n_{i=1}M_i\Delta x_i \qquad L(f,P) = \sum^n_{i=1}m_i\Delta x_i

as the Upper sum and Lower sum respectively.

The upper (lower) sums are defined as the sum of the product of the supremum (infimum) of each subinterval multiplied by the length of the subinterval.

Because we assume f to be bounded function, there exists a lower and upper bound of the function on the interval [a,b]. This also implies the existence of a supremum and infimum, so we know our upper and lower sums exist.

Call these bounds m,M respectively (No relation to m_i,M_i). For any partition P of [a,b] we have m(b-a) \leq L(f,p) \leq U(f,p) \leq M(b-a).

Thus we know the upper and lower sums for f exist, and form a bounded set.

Theorem 4.1.1

Let f be a bounded function on [a,b]. If P and Q are partitions of [a,b], and Q is a refinement of P, then

L(f,P) \leq L(f,Q) \leq U(f,Q) \leq U(f,P).

Direct Proof

The middle inequality L(f,Q) \leq U(f,Q) above follows from our definitions. To prove L(f,P) \leq L(f,Q), suppose P = \{x_0,\ldots,x_k\}, and consider the partition P* formed by adding x* to P, where x_{k-1} < x* < x_k for some k \in \{1,\ldots,n\}. Let

\begin{align*} t_1 &= \inf \{f(x) \colon x \in [x_{k-1},x*]\}\\ t_2 &= \inf \{f(x) \colon x \in [x*, x_k]\} \end{align*}

Then t_1 \geq m_k and t_2 \geq m_k, where m_k = \inf \{f(x) \colon x \in [x_{k-1},x_k]\} as defined previously. All of the terms in L(f,P*) and L(f,P) are the same except for those over the interval [x_{k-1},x_k]. Thus we have

\begin{align*} L(f,P*) - L(f,P) &= [t_1(x* - x_{k-1}) + t_2(x_k - x*)] - [m_k(x_k - x_{k-1})]\\ &= (t_1 - m_k)(x* - x_{k-1}) + (t_2 - m_k)(x_k - x*). \end{align*}

This final sum is positive since all of the terms are positive. Thus L(f,P) \leq L(f,P*). If our partition Q contains r more points than P, we apply this argument r times to obtain L(f,P) \leq L(f,Q).

The proof of U(f,Q) \leq U(f,P) is similar.

Theorem 4.1.2

Let f be a bounded function on [a,b]. If P and Q are partitions of [a,b], then L(f,P) \leq U(f,Q).

Direct Proof

If P and Q are equal the proof is trivial, so we need only consider the case where P and Q are seperate partitions. Let R be the partition given by P \cup Q. R is a refinement of both P and Q. By Theorem 4.1.1 we have

\begin{align*} L(f,P) \leq L(f,R) &\leq U(f,R) \leq U(f,P)\\ L(f,Q) \leq L(f,R) &\leq U(f,R) \leq U(f,Q)\\ \end{align*}

Thus L(f,P) \leq U(f,Q) for all partitions P and Q.

Riemann Integral

Definition

Let f be a bounded function defined on [a,b] and P be a partition on [a,b]. Then

U(f) = \inf \{U(f,P)\} \qquad L(f) = \sup \{L(f,P)\}

are called the upper integral and lower integral of f on [a,b], respectively.

Definition

If L(f) = U(f), we say that f is Riemann integrable on [a,b], and notate the value with

\int^a_b f \quad \text{ or } \quad \int^a_b f(x) \, dx.

This is known as the Riemann integral of f on [a,b].

If a function is Riemann integrable, then the value of the integral on [a,b] will be equal to the signed area between the function and the x-axis.

Corollary 4.1.3

Let f be a bounded function on [a,b]. Then L(f) \leq U(f).

Direct Proof

If P and Q are partitions of [a,b], we have that L(f,P) \leq U(f,Q) by Theorem 4.1.2. Then it follows that U(f,Q) is an upper bound of the set defined by

S = \{L(f,P) \colon P \text{ partitions } [a,b]\}.

Thus U(f,Q) \geq \sup S. That is, L(f) \leq U(f,Q) for all partitions Q of [a,b]. Hence

L(f) \leq \inf \{U(f,Q) \colon Q \text{ partitions } [a,b]\} = U(f).

Essentially this Corollary states that every lower sum will be less than any upper sum.

Not every function is going to be Riemann integrable.

Example (Non-integrable functions)

Let f \colon [0,1] \to \mathbb{R} be defined

g(x) = \begin{cases} 1, & \text{if $x$ is rational}\\ 0, & \text{if $x$ is irrational.} \end{cases}

Let P = \{x_0,\ldots,x_n\} be any partition of [0,1]. Since each subinterval [x_{i-1},x_i] contains both rational and irrational points, we have M_i = 1 and m_i = 0 for all i \in \{1,\dots,n\}. Therefore

U(f,P) = \sum^n_{i=1}(1)\Delta x_i = 1 \qquad L(f,P) = \sum^n_{i=1}(0)\Delta x_i = 0.

Since the lower and upper integrals are different for every partition P, f is non-integrable on [0,1]. This special function f is known as the Dirichlet function.

Theorem 4.1.4 (Riemann integrablility Criterion)

Let f be a bounded function on [a,b]. Then f is integrable if and only if for each \varepsilon > 0 there exists a partition P of [a,b] such that

U(f,P) - L(f,P) < \varepsilon.

Direct Proof

Forward direction

Suppose f is integrable, meaning L(f) = U(f). Given any \varepsilon > 0, there exists a partition P_1 of [a,b] such that

L(f,P_1) > L(f) - \frac{\varepsilon}{2}.

This follows since L(f) is a supremum. Similarly, there exists a partition P_2 of [a,b] such that

U(f,P_2) < U(f) + \frac{\varepsilon}{2}.

Let P = P_1 \cup P_2. Then by Theorem 4.1.1 we have

\begin{align*} U(f,P) - L(f,P) &\leq U(f,P_2) - L(f,P_1)\\ &< \left[U(f) + \frac{\varepsilon}{2}\right] - \left[L(f) - \frac{\varepsilon}{2}\right]\\ &= U(f) - L(f) + \varepsilon = \varepsilon. \end{align*}

Therefore f being integrable implies there exists P such that U(f,P) - L(f,P) < \varepsilon.

Reverse direction

Now, suppose we have P such that U(f,P) - L(f,P) < \varepsilon. Then

U(f) \leq U(f,P) < L(f,P) + \varepsilon \leq L(f) + \varepsilon.

Since \varepsilon > 0 is arbitrary, we have U(f) \leq L(f). By Theorem 4.1.3 we have that L(f) \leq U(f). Thus L(f) = U(f), and f is Riemann integrable.