# Riemann integral/Above and below/Section

## Definition

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

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

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

of ${\displaystyle {}f}$ exist and coincide.

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

## Definition

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

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

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

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

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

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

## Lemma

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

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

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

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

### Proof

${\displaystyle \Box }$

## Example

We consider the function

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

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

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

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

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

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

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

## Lemma

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

${\displaystyle f\colon I\longrightarrow \mathbb {R} }$
be a function. Then the following statements are equivalent.
1. The function ${\displaystyle {}f}$ is Riemann-integrable.
2. There exists a partition ${\displaystyle {}a=a_{0}, such that the restrictions ${\displaystyle {}f_{i}:=f|_{[a_{i-1},a_{i}]}}$ are Riemann-integrable.
3. For every partition ${\displaystyle {}a=a_{0}, the restrictions ${\displaystyle {}f_{i}:=f|_{[a_{i-1},a_{i}]}}$ are Riemann-integrable.
In this situation, the equation
${\displaystyle {}\int _{a}^{b}f(t)\,dt=\sum _{i=1}^{n}\int _{a_{i-1}}^{a_{i}}f_{i}(t)\,dt\,}$

holds.

### Proof

${\displaystyle \Box }$

## Definition

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

Riemann-integrable.

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