Real Analysis/Riemann integration

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←Applications of Derivatives Real Analysis
Riemann integration		 Fundamental Theorem of Calculus→

Contents

[edit] Definition

Riemann integration is the formulation of integration most people think of if they ever think about integration. It is the only type of integration considered in most calculus classes; many other forms of integration, notably Lebesgue integrals, are extensions of Riemann integrals to larger classes of functions. The Riemann integral was developed by Bernhard Riemann in 1854 and was, when invented, the first rigorous definition of integration applicable to not necessarily continuous functions.

We will first define some preliminary ideas.

[edit] Partitions

[edit] Definition

Let a,b\in\mathbb{R}

A Partition \mathcal{P} is defined as the ordered n-tuple of real numbers \mathcal{P}=(x_1,x_2,\ldots ,x_n) such that a<x_1<x_2<\ldots <x_n<b

[edit] Norm of a Partition

Let \mathcal{P} be a partition given by \mathcal{P}=(x_1,x_2,\ldots ,x_n)

Then, the Norm (or the "mesh") of \mathcal{P} is defined as \|\mathcal{P}\| =\sup \{x_{k+1}-x_k|1\leq k\leq n-1\}

[edit] Tagged Partition

Let \mathcal{P}=(x_1,x_2,\ldots ,x_n) be a partition

A Tagged Partition \mathcal{\dot{P}} is defined as the set of ordered pairs \mathcal{\dot{P}}=\{([x_{i-1},x_i],t_i)\}_{i=1}^n such that xi − 1 < ti < xi. The points ti are called Tags.

[edit] Riemann Integration

Riemann sum of a function

[edit] Riemann Sums

Let f:[a,b]\to\mathbb{R}

Let \mathcal{\dot{P}}=\{([x_{i-1},x_i],t_i)\}_{i=1}^n be a tagged partition of [a,b]

The Riemann Sum of f over [a,b] with respect to \mathcal{\dot{P}} is given by

S(f,\mathcal{\dot{P}})=\displaystyle\sum_{i=1}^n f(t_i)(x_i-x_{i-1})

[edit] Riemann Integral

Let f:[a,b]\to\mathbb{R}

Let L\in\mathbb{R}

We say that f is Integrable on [a,b] if and only if for every \varepsilon>0, there exists δ > 0 such that for every partition \mathcal{\dot{P}} satisfying \|\mathcal{\dot{P}}\|<\delta, we have that |S(f,\mathcal{\dot{P}})-L|<\varepsilon

L is said to be the integral of f over [a,b], and is written as

L=\int_a^b f(x)dx or as L=\int_a^b f

[edit] Properties

[edit] Theorem (Uniqueness)

Let f:[a,b]\to\mathbb{R} be integrable on [a,b]

Then the integral L of f is unique

[edit] Proof

Assume, if possible that L_1\neq L_2 are both integrals of f over [a,b]. Consider \varepsilon=\tfrac{|L_1-L_2|}{2}

As L1,L2 are integrals, there exist δ1,δ2 > 0 such that |S(f,\mathcal{\dot{P}})-L_1|<\varepsilon for all \mathcal{\dot{P}} that satisfy \|\mathcal{\dot{P}}\|<\delta_1 and |S(f,\mathcal{\dot{P}})-L_2|<\varepsilon for all \mathcal{\dot{P}} that satisfy \|\mathcal{\dot{P}}\|<\delta_2

Let \delta=\inf\{\delta_1,\delta_2\}. Hence, if \mathcal{\dot{P}} is a partition satisfying \|\mathcal{\dot{P}}\|<\delta, then we have |S(f,\mathcal{\dot{P}})-L_1|<\varepsilon and that |S(f,\mathcal{\dot{P}})-L_2|<\varepsilon

That is, |L_1-L_2|<2\varepsilon <|L_1-L_2|, which is an obvious contradiction. Hence the integral L of f is unique.


We now state (without proof) two seemingly obvious properties of the integral.

[edit] Theorem

Let f,g:[a,b]\to\mathbb{R} be integrable and let c\in (a,b)

Then:

(i)\int_a^b f+\int_a^b g=\int_a^b f+g

(ii)\int_a^c f+\int_c^b f=\int_a^b f

[edit] Theorem (Boundedness Theorem)

Let f:[a,b]\to\mathbb{R} be Riemann integrable. Then, f is bounded over [a,b]

[edit] Proof

Assume if possible that f is unbounded. For every n\in\mathbb{N} divide the interval [a,b] into n parts. Hence, for every n\in\mathbb{N}, f is unbounded on at least one of these n parts. Call it In.

Now, let \varepsilon>0 be given. Consider an arbitrary δ > 0. Let \mathcal{\dot{P}} be a tagged partition such that \|\mathcal{\dot{P}}\|<\delta and (I_n,t_n)\in\mathcal{\dot{P}}, where tn is taken so as to satisfy |f(t_n)|>n\varepsilon.

Thus we have that |S(f,\mathcal{\dot{P}})-L|>\varepsilon. But as δ > 0 is arbitrary, we have a contradiction to the fact that f is Riemann integrable.

Hence, f is bounded.

[edit] Integrability

We now study classes of Riemann integrable functions. The first "constraint" on Riemann integrable functions is provided by the Cauchy Integrability Criterion.

[edit] Theorem (Cauchy Criterion)

Let f:[a,b]\to\mathbb{R}

Then,

(i)f is Riemann integrable on [a,b] if and only if

(ii) For every \varepsilon>0, there exists δ > 0 such that if \mathcal{\dot{P}},\mathcal{\dot{Q}} are two partitions satisfying \|\mathcal{\dot{P}}\|,\|\mathcal{\dot{Q}}\|<\delta then |S(f,\mathcal{\dot{P}})-S(f,\mathcal{\dot{Q}})|<\varepsilon

[edit] Proof

(\Rightarrow)Let \int_a^b f=L and let \varepsilon>0 be given.

Then, there exists δ > 0 such that for every partition \mathcal{\dot{P}} satisfying \|\mathcal{\dot{P}}\|<\delta,we have |S(f,\mathcal{\dot{P}})-L|<\tfrac{\varepsilon}{2}

Now, let partitions \mathcal{\dot{P}},\mathcal{\dot{Q}} be such that \|\mathcal{\dot{P}}\|,\|\mathcal{\dot{Q}}\|<\delta.

Thus we have that |S(f,\mathcal{\dot{P}})-L|,|S(f,\mathcal{\dot{Q}})-L|<\tfrac{\varepsilon}{2}, that is |S(f,\mathcal{\dot{P}})-S(f,\mathcal{\dot{Q}})|<\varepsilon

(\Leftarrow) For every n\in\mathbb{N}, consider δn > 0 such that for all partitions \mathcal{\dot{P}},\mathcal{\dot{Q}} satisfying \|\mathcal{\dot{P}}\|,\|\mathcal{\dot{Q}}\|<\delta_n, we have |S(f,\mathcal{\dot{P}})-S(f,\mathcal{\dot{Q}})|<\tfrac{1}{n}.

Without loss of generality, we can assume that δm > δn when m < n. For every δn, let \mathcal{\dot{P}}_n be a partition such that \|\mathcal{\dot{P}}_n\|<\delta_n

The sequence a_n=S(f,\mathcal{\dot{P}}_n) is a Cauchy sequence, and hence it has a limit L\in\mathbb{R}.

Now, for every \varepsilon>0, we have a δ > 0 such that \|\mathcal{\dot{P}}\|<\delta implies |S(f,\mathcal{\dot{P}}-L)|<\varepsilon.

Thus \int_a^b f=L

[edit] Theorem (Squeeze Theorem)

Let f:[a,b]\to\mathbb{R}

Then,

(i) f is Riemann integrable on [a,b] if and only if

(ii) For every \varepsilon>0, there exist Riemann integrable functions \alpha_{\varepsilon},\omega_{\varepsilon}:[a,b]\to\mathbb{R} such that

\alpha_{\varepsilon}(x)\leq f(x)\leq\omega_{\varepsilon}(x) for all x\in[a,b] and

\int_a^b (\omega_{\varepsilon}-\alpha_{\varepsilon})<\varepsilon

[edit] Proof

(\Rightarrow)Take \alpha_{\varepsilon}(x)=f(x)=\omega_{\varepsilon}(x). It is easy to see that \int_a^b (\omega_{\varepsilon}-\alpha_{\varepsilon})<\varepsilon

(\Leftarrow)Let n\in\mathbb{N}. Then, there exist functions αn,ωn such that \int_a^b (\omega_n-\alpha_n)<\tfrac{1}{n}. Further, if \int_a^b \alpha_n=A_n and \int_a^b \omega_n=Z_n, then there exist δ1,δ2 > 0 such that if a partition \mathcal{\dot{P}} satisfies \|\mathcal{\dot{P}}\|<\delta_1 then |S(\alpha_n,\mathcal{\dot{P}})-A_n|<\tfrac{1}{n} and \|\mathcal{\dot{P}}\|<\delta_2 then |S(\omega_n,\mathcal{\dot{P}})-Z_n|<\tfrac{1}{n}

Now let \mathcal{\dot{P}}_n be an interval satisfying \|\mathcal{\dot{P}}_n\|<\inf \{\delta_1,\delta_2\}.

Now, we can easily see that |S(f,\mathcal{\dot{P}}_n)-S(f,\mathcal{\dot{P}}_{n-1})|<\tfrac{1}{n}. Hence, S(f,\mathcal{\dot{P}}_n) is a Cauchy sequence, with a limit L\in\mathbb{R}, and as in the previous proof, we can show that \int_a^b f=L

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