Riemann Integral

Integration of Real-valued Functions

Motivation

Given an interval $[a, b]$ and some function $f: [a, b] \rightarrow \mathbb{R}$, we want to create some rule, that computes the area under the curve $y = f(x)$ between $x = a$ and $x = b$.

This rule, which assigns real numbers to functions, can be itself represented as a function $\varphi_{[a, b]} : \mathcal{F} \big( [a, b] \big) \rightarrow \mathbb{R}$. Here, $\mathcal{F} \big( [a, b] \big)$, the domain of $\varphi$, refers to a collection of functions defined on $[a, b]$ such that the area under the curve $y = f(x)$ between $x = a$ and $x = b$ is well-defined according to the rule.

The integral of a function $f \in \mathcal{F} \big( [a, b] \big)$ is defined as the value obtained under the rule $\varphi_{[a, b]}$. To represent this value, we conventionally use the notation:

Similarly, when considering another interval $[s, t]$, we can establish a similar rule $\varphi_{[s, t]}$ with domain $\mathcal{F} \big( [s, t] \big)$. The integral of $f$ over $[s, t]$ is denoted as:

In the subsequent discussion, we will employ the notation on the left-hand side exclusively.

The integral should adhere to our geometric understanding of areas and satisfy various properties. First and foremost, the integral should accurately compute the area of rectangles.

Property (Area of Rectangle). If $\exists c \in \mathbb{R}$, s.t., $f(x) = c$ for each $x \in [a, b]$, then $f \in \mathcal{F} \big( [a, b] \big)$ and $\int_a^b f \, dx = c \cdot (b - a)$. .

Secondly, the integrals should demonstrate linearity.

Property (Linearity).

1. If $f \in \mathcal{F} \big( [a, b] \big)$ and $c \in \mathbb{R}$, then $c \cdot f \in \mathcal{F} \big( [a, b] \big)$ and

$$\int _{a}^{b} c \cdot f \, dx = c \cdot \int_{a}^{b} f \, dx.$$

2. If $f_1, f_2 \in \mathcal{F} \big( [a, b] \big)$, then $f_1 + f_2 \in \mathcal{F} \big( [a, b] \big)$, and

$$\int _{a}^{b} f_1 + f_2 \, dx = \int_{a}^{b} f_1 \, dx + \int_{a}^{b} f_2 \, dx.$$

3. If $f \in \mathcal{F} \big( [a, b] \big)$ and $c \in (a, b)$, then $f \in \mathcal{F} \big( [a, c] \big)$ and $f \in \mathcal{F} \big( [c, b] \big)$, and

$$\int _{a}^{b} f \, dx = \int _{a}^{c} f \, dx + \int _{c}^{b} f \, dx.$$

The integrals should allow for meaningful comparisons.

Property (Monotonicity).

1. If $f_1, f_2 \in \mathcal{F} \big( [a, b] \big)$ and $f_1 \le f_2$, then

$$\int_{a}^{b} f_1 \, dx \le \int_{a}^{b} f_2 \, dx.$$

2. If $f \in \mathcal{F} \big( [a, b] \big)$ and $|f| \le M$, then

$$\left| \int_{a}^{b} f \, dx \right| \le M \cdot (b - a).$$

So far, it is straightforward to specify a family of functions, and a rule that assigning real numbers to them,satisfying the aforementioned properties. For instance, we can define $\mathcal{F}\big( [a, b] \big)$ as the set of step functions on $[a, b]$. Each step function $f$ can be expressed as:

where $E_i, i \in [n]$, are sub-intervals, forming a partition of $[a, b]$. Further, we can define the integral as

where $|E_i|$ is the length of $E_i$.

However, it is also desirable for the constructed integrals to satisfy convergence properties.

Property (Convergence).

1. If $f_i \in \mathcal{F} \big( [a, b] \big)$ for each $i \in \mathbb{N}^+$ and $f_i \rightarrow f$ under some metric, then $f \in \mathcal{F} \big( [a, b] \big)$ and

$$\lim_{n \rightarrow \infty} \int_{a}^{b} f_i \, dx = \int_{a}^{b} f \, dx.$$

Unfortunately, the previous construction does not satisfy convergence properties. Additional considerations and refinements are required in order to establish an integral that fulfills the desired convergence properties.

Construction

Bernhard Riemann presented the following construction of integral.

Let $[a, b]$ be a given interval. $P = \{ x_0 , x_1 , \ldots, x_n \}$ is a partition of $[a, b]$, if

For each $i \in [n]$, let $\Delta x_i \doteq x_i, - x_{i - 1}$.

Assume $f: [a, b] \rightarrow \mathbb{R}$ is a bounded function. Associated with each partition $P$, define

Further, let

Finally, consider

where the inf and sup are taken over all partitions $P$ of $[a, b]$.

Definition $f: [a, b] \rightarrow \mathbb{R}$ is considered Riemann integrable, if

$$\overline{\int _{a}^{b}} f \, dx = \underline{\int_{a}^{b}} f \, dx.$$

In this cases, we define the integral of $f$ over the interval $[a, b]$ as:

$$\int _{a}^{b} f \, dx \doteq \overline{\int _{a}^{b}} f \, dx = \underline{\int_{a}^{b}} f \, dx.$$

We denote the collection of functions that are Riemann integrable on $[a, b]$ as $\mathfrak{R}\big( [a, b] \big)$.

Remark. It is reasonable to assume that $f$ is bounded on the closed interval $[a, b]$, in order that it is Riemann integrable. If $f$ is unbounded, for each partition $P$, there exists some $i \in [n]$, s.t., $M_i - m_i = \infty$. Hence it is impossible that $\overline{\int _{a}^{b}} f \, dx = \underline{\int_{a}^{b}} f \, dx$.

Properties

It is straightforward to see the proposed construction of integral captures the area of rectangles, and demonstrates linearity and monotonicity.

Theorem. Assume that for each $n \in \mathbb{N}^+$, $f_n \in \mathfrak{R} \big( [a, b] \big)$, and that $f_n \rightarrow f$ uniformly on $[a, b]$ (i.e., $\sup_{x \in [a, b]} |f_n(x) - f(x)| \rightarrow 0$). Then $f\in \mathfrak{R} \big( [a, b] \big)$, and

$$\int_a^b f \, dx = \lim_{n \rightarrow \infty} \int_a^b f_n \, dx.$$

(The existence of the limit is part of the conclusion.)

Proof. Choose $\varepsilon > 0$. For large enough $n$, we have $|f_n - f| \le \varepsilon$.

Let $P = \{ x_0 , x_1 , \ldots, x_n \}$ be a partition of $[a, b]$.

Taking sup and inf over all possible partitions of $P$ on $[a, b]$, and given the Riemann integrability of $f_n$, we have

Since $\varepsilon$ is chosen arbitrarily, we have $\underline{\int_{a}^{b}} f \, dx = \overline{\int _{a}^{b}} f \, dx$ and therefore $f \in \mathfrak{R} \big( [a, b] \big)$.

Finally,

implies that $\lim_{n \rightarrow \infty} \int_a^b f_n \, dx$ exists, and equals $\int_{a}^{b} f \, dx$.

$\square$

Integration and Differentiation

Theorem. Suppose that $f \in \mathfrak{R} \big( [a, b] \big)$. Define

$$F(t) \doteq \int_a^t f \, dx, \quad \forall t \in [a, b].$$

Then $F$ is continuous on $[a, b]$; furthermore, if $f$ is continuous at a point $t \in [a, b]$, then $F$ is differentiable at $t$, and $F'(t) = f(t)$.

Proof. Since $f\in \mathfrak{R} \big( [a, b] \big)$, it is bounded. Let $M \doteq \max_{x \in [a, b]} |f(x)|$. It follows that

which proves uniform continuity of $F$.

With out lose of generality, assume that $t \in (a, b)$. Since $f$ is continuous at $t$, $\exists \delta > 0$, s.t., $[t - \delta, t + \delta] \subseteq [a, b]$ and $\forall s \in [t - \delta, t + \delta]$, it holds that $|f(s) - f(t)| \le \epsilon$. It follows that for all $\ell, r \in [t - \delta, t + \delta]$,

$\square$

Theorem (The fundamental theorem of calculus). Suppose that $f \in \mathfrak{R} \big( [a, b] \big)$ and that $F$ are differentiable function on $[a, b]$ such that $F' = f$, then

$$\int_a^b f \, dx = F(b) - F(a).$$

Proof. Let $P = \{ x_0 , x_1 , \ldots, x_n \}$ be a partition of $[a, b]$. For each $i \in [n]$, by mean value theorem, $\exists t_i \in [x_{i - 1}, x_i]$, s.t.,

Hence,

Taking sup and inf over all partitions $P$ finishes the proof.

$\square$

Corollary (Integration by parts). If $f_1, f_2\in \mathfrak{R} \big( [a, b] \big)$, and $F_1$ and $F_2$ are differentiable function on $[a, b]$ such that $F_1' = f_1$ and $F_2' = f_2$, then

$$\int_a^b F_1 \cdot f_2 \, dx = F_1(b) \cdot F_2(b) - F_1(a) \cdot F_2(a) - \int_a^b F_2 \cdot f_1 \, dx.$$

Existence

In this section, we delve into the functions are encompassed within $\mathfrak{R} \big( [a, b] \big)$.

Theorem.

1. If $f$ is continuous on $[a, b]$ then $f\in \mathfrak{R} \big( [a, b] \big)$.
2. If $f$ is monotonic on $[a, b]$, then $f\in \mathfrak{R} \big( [a, b] \big)$.
3. If $f$ is bounded on $[a, b]$ and has only finitely many points of discontinuity on $[a, b]$, then $f\in \mathfrak{R} \big( [a, b] \big)$.

Theorem. Suppose

1. $f \in \mathfrak{R} \big( [a, b] \big)$, and $f([a, b]) \subseteq [m, M]$,
2. $g: [m, M] \rightarrow \mathbb{R}$ is continuous, and
3. $h \doteq g \circ f$.

Then $h \in \mathfrak{R} \big( [a, b] \big)$.

Note that if $f \in \mathfrak{R} \big( [a, b] \big)$, then it is bounded. Therefore, we can always find a closed interval $[m, M] \subseteq \mathbb{R}$ such that $f([a, b]) \subseteq [m, M]$. The only requirement of the theorem is that, the continuous function $g$ is defined over some of such intervals.

Proof. Consider a given $\varepsilon > 0$.

Since $g$ is continuous on $[m, M]$, there exists $K \doteq \sup_{t \in [m, M]} | g(t) |$. Further, it is uniformly continuous on $[m, M]$. Hence,

Since $f \in \mathfrak{R} \big( [a, b] \big)$, there is a partition $P = \{ x_0 , x_1 , \ldots, x_n \}$ of $[a, b]$ such that

Let

Clearly it holds that $|M_i^*| \le K$ and $|m_i^*| \le K$. Define

Then we have

• $\forall i \in A$, $M_i^* - m_i^* \le \varepsilon$, and
• $\sum_{i \in B} \Delta x_i \le \delta$ since

It follows that

$\square$

Corollary. If $f_1, f_2 \in \mathfrak{R} \big( [a, b] \big)$, then

1. $f_1 \cdot f_2\in \mathfrak{R} \big( [a, b] \big)$;
2. $|f_1|\in \mathfrak{R} \big( [a, b] \big)$ and

$$\left| \int _{a}^{b} f_1 \, dx \right| \le \int_{a}^{b} | f_1 | \, dx.$$

Proof. Since the square function is continuous on $\mathbb{R}$, it follows that $(f_1 + f_2)^2, (f_1 - f_2)^2 \in \mathfrak{R} \big( [a, b] \big)$. Claim 1 follows from that $4 \cdot f_1 \cdot f_2 = (f_1 + f_2)^2 - (f_1 - f_2)^2$.

Since the absolute function is continuous on $\mathbb{R}$, it follows that $|f_1|\in \mathfrak{R} \big( [a, b] \big)$. The inequality follows from monotonicity of Riemann integration.

$\square$

Integration of Vector-valued Functions

Let $\vec{f} \doteq (f_1, \ldots , f_k): [a, b] \rightarrow \mathbb{R}^k$ vector-valued function. We call it Riemann integrable if for each $i \in [k]$, $f_i \in \mathfrak{R} \big( [a, b] \big)$. In this case, we write

Properties

Theorem. If $\vec{f} \doteq (f_1, \ldots , f_k): [a, b] \rightarrow \mathbb{R}^k$ is Riemann integrable, then

$$\lVert \vec{f} \rVert_2 \doteq \sqrt{f_1^2 + \cdots + f_k^2} \in \mathfrak{R} \big( [a, b] \big).$$

Further it holds that

$$\left\lVert \int_a^b \vec{f} \, dx \right\rVert_2 \le \int_a^b \lVert \vec{f} \rVert_2 \, dx.$$

Proof. Given the Riemann integrability of $f_i$ on $[a, b]$ and the continuity of the square function on $\mathbb{R}$, we can deduce that $f_i^2 \in \mathfrak{R} \big( [a, b] \big)$. Thus, $\lVert \vec{f} \rVert_2^2 \in \mathfrak{R} \big( [a, b] \big)$.

Moreover, considering the Riemann integrability of each $f_i$ on the interval $[a, b]$, we can infer that they are bounded. Hence, we can assume an upper bound $M$ for each $f_i^2$.

Furthermore, since the square root function is continuous on the interval $[0, M]$, we can conclude that $\lVert \vec{f} \rVert_2 \in \mathfrak{R} \big( [a, b] \big)$.

Let $\vec{y} = (y_1, \ldots, y_k)$ be a shorthand for $\int_a^b \vec{f} \, dx$. Then

$\square$

Reference

[1] W. Rudin, Principles of mathematical analysis, 3d ed. in International series in pure and applied mathematics. New York: McGraw-Hill, 1976.