Riemann integral/Above and below/Section

It might by historically more adequate to call this Darboux-integrable.

A step function from above and one from below. The green area is a step integral from below and the yellow area (partly covered) is a step function from above.

Definition

Let ${}I$ denote a compact interval and let

$f\colon I\longrightarrow \mathbb {R}$

denote a function. Then ${}f$ is called Riemann-integrable if the upper integral and the lower integral

of ${}f$ exist and coincide.

Definition

Let ${}I=[a,b]$ denote a compact interval. For a Riemann-integrable function

$f\colon I=[a,b]\longrightarrow \mathbb {R} ,t\longmapsto f(t),$

we call the upper integral of ${}f$ (which by definition coincides with the lower integral) the definite integral of ${}f$ over ${}I$. It is denoted by

$\int _{a}^{b}f(t)\,dt{\text{ or by }}\int _{I}^{}f(t)\,dt.$

The computation of such integrals is called to integrate. Don't think too much about the symbol ${}dt$. It expresses that we want to integrate with respect to this variable. The name of the variable is not relevant, we have

${}\int _{a}^{b}f(t)\,dt=\int _{a}^{b}f(x)\,dx\,.$

Lemma

Let ${}I$ denote a compact interval, and let

$f\colon I\longrightarrow \mathbb {R}$

denote a function. Suppose that there exists a sequence of lower step functions ${}{\left(s_{n}\right)}_{n\in \mathbb {N} }$ with ${}s_{n}\leq f$ and a sequence of upper step functions ${}{\left(t_{n}\right)}_{n\in \mathbb {N} }$ with ${}t_{n}\geq f$. Suppose furthermore that the corresponding sequences of step integrals converge to the same real number. Then ${}f$ is Riemann-integrable, and the definite integral equals this limit, so

${}\lim _{n\rightarrow \infty }\int _{a}^{b}s_{n}(x)\,dx=\int _{a}^{b}f(x)\,dx=\lim _{n\rightarrow \infty }\int _{a}^{b}t_{n}(x)\,dx\,.$

Proof

See exercise.

$\Box$

Example

We consider the function

$f\colon [0,1]\longrightarrow \mathbb {R} ,t\longmapsto t^{2},$

which is strictly increasing in this interval. Hence, for a subinterval ${}[a,b]\subseteq [0,1]$, the value ${}f(a)$ is the minimum, and ${}f(b)$ is the maximum of the function on this subinterval. Let ${}n$ be a positive natural number. We partition the interval ${}[0,1]$ into the ${}n$ subintervals ${}\left[i{\frac {1}{n}},(i+1){\frac {1}{n}}\right]$, ${}i=0,\ldots ,n-1$, of length ${}{\frac {1}{n}}$. The step integral for the corresponding lower step function is

${}\sum _{i=0}^{n-1}{\frac {1}{n}}{\left(i{\frac {1}{n}}\right)}^{2}={\frac {1}{n^{3}}}\sum _{i=0}^{n-1}i^{2}={\frac {1}{n^{3}}}{\left({\frac {1}{3}}n^{3}-{\frac {1}{2}}n^{2}+{\frac {1}{6}}n\right)}={\frac {1}{3}}-{\frac {1}{2n}}+{\frac {1}{6n^{2}}}\,$

(see exercise for the formula for the sum of the squares). Since the sequences ${}{\left(1/2n\right)}_{n\in \mathbb {N} }$ and ${}{\left(1/6n^{2}\right)}_{n\in \mathbb {N} }$ converge to ${}0$, the limit for ${}n\rightarrow \infty$ of these step integrals equals ${}{\frac {1}{3}}$. The step integral for the corresponding step function from above is

${}\sum _{i=0}^{n-1}{\frac {1}{n}}{\left((i+1){\frac {1}{n}}\right)}^{2}={\frac {1}{n^{3}}}\sum _{i=0}^{n-1}(i+1)^{2}={\frac {1}{n^{3}}}\sum _{j=1}^{n}j^{2}={\frac {1}{n^{3}}}{\left({\frac {1}{3}}n^{3}+{\frac {1}{2}}n^{2}+{\frac {1}{6}}n\right)}={\frac {1}{3}}+{\frac {1}{2n}}+{\frac {1}{6n^{2}}}\,.$

The limit of this sequence is again ${}{\frac {1}{3}}$. By fact, the upper integral and the lower integral coincide, hence the function is Riemann-integrable, and for the definite integral we get

${}\int _{0}^{1}t^{2}\,dt={\frac {1}{3}}\,.$

Lemma

Let ${}I=[a,b]$ be a compact interval, and let

$f\colon I\longrightarrow \mathbb {R}$

be a function. Then the following statements are equivalent.

1. The function ${}f$ is Riemann-integrable.
2. There exists a partition ${}a=a_{0}<a_{1}<\cdots <a_{n}=b$, such that the restrictions ${}f_{i}:=f|_{[a_{i-1},a_{i}]}$ are Riemann-integrable.
3. For every partition ${}a=a_{0}<a_{1}<\cdots <a_{n}=b$, the restrictions ${}f_{i}:=f|_{[a_{i-1},a_{i}]}$ are Riemann-integrable.

In this situation, the equation

${}\int _{a}^{b}f(t)\,dt=\sum _{i=1}^{n}\int _{a_{i-1}}^{a_{i}}f_{i}(t)\,dt\,$

holds.

Proof

See exercise.

$\Box$

Definition

Let ${}f\colon I\rightarrow \mathbb {R}$ be a function on a real interval. Then ${}f$ is called Riemann-integrable, if the restriction of ${}f$ to every compact interval ${}[a,b]\subseteq I$ is

Riemann-integrable.

Due to this lemma, both definitions coincide for a compact interval ${}[a,b]$. The integrability of a function ${}f\colon \mathbb {R} \rightarrow \mathbb {R}$ does not mean that ${}\int _{\mathbb {R} }f(x)dx$ has a meaning or exists.