Riemann integrable
Riemann integrable
A bounded function \[f:\left[ a,b \right]\to\mathbb{R}\] is Riemann integrable on \[\left[ a,b \right]\] if its upper Riemann integral \[U(f)\] and lower Riemann integral \[L(f)\] are essentially equal. Then, Riemann integral of \[f\] on \[\left[ a,b \right]\] can be written as something we're all familiar with, \[U(f)=L(f)=\int_{a}^{b}f(x)\,dx\].
This reason why when \[U(f)=L(f)\], then \[f\] is Riemann integrable, is that \[U(f)-L(f)\] is the difference between the infimum of the upper Riemann sum and supremum of the lower Riemann sum. Since \[U(f)\] and \[L(f)\] is defined on every possible partition (\[\Pi\]), it is guaranteed that we can find a partition \[P\] with sufficient amount of points (a partition fine enough, thus making the sums increasingly accurate), such that the difference will be as small as we want (i.e. \[\epsilon>0\]).
Formally, \[f:\left[ a,b \right]\to\mathbb{R}\] is Riemann integrable on \[\left[ a,b \right]\] if and only if any \[\epsilon>0\] we can find a partition \[P\] of \[\left[ a,b \right]\] such that \[U(f;P)-L(f;P)<\epsilon\].
Example
Define \[f:\left[ 0,1 \right]\to\mathbb{R}\] by \[f(x)=\begin{cases} \frac{1}{x} & 0<x\le1 \\ 0 & x=0 \end{cases}\]. Then \[\int_{0}^{1}\frac{1}{x}\,dx\] is not Riemann integrable (undefined) as \[f\] is unbounded. In fact, if we define \[P\] as endpoints, then \[\sup_{\left[ x_{0},x_{1} \right]}f=\infty\] (as \[\lim_{x\to0+}f(x)=\infty\]), it follows that \[U(f;P)\] is also not well-defined.
Similarly, \[\int_{1}^{\infty}\frac{1}{x}\,dx\] isn't Riemann integrable either as a partition of \[[1,\infty)\] into a finite number of intervals would include at least one unbounded interval, making the Riemann sum not well-defined. If it's partitioned into an infinite number of bounded intervals (e.g. \[I_{k}=\left[ k,k+1 \right],k\in\mathbb{N}\]), then it would give us an infinite series rather than a finite sum.
Now, an example of something that's Riemann integrable is \[f(x)=1\] on \[\left[ 0,1 \right]\], and \[\int_{0}^{1}1\,dx=1\]. To prove that, let \[P\] be any partition of \[\left[ 0,1 \right]\] with endpoints \[\left\{ 0,x_{1},x_{2},\dots,x_{n-1},1 \right\}\]. Since \[f\] is constant, \[M_{k}=\sup_{I_{k}}f=1\] and \[m_{k}=\inf_{I_{k}}f=1\] for \[k\in\mathbb{N}\]. Therefore, \[U(f;P)=L(f;P)=\sum_{k=1}^{n}\left( x_{k}-x_{k-1} \right)=x_{n}-x_{0}=1\]. Thus, every upper and lower sum is equal to 1, which implies that \[U(f)=\inf_{P\in\Pi}U(f;P)=1\] and \[L(f)=\sup_{P\in\Pi}L(f;P)=1\]. Since \[U(f)=L(f)\], we say that \[\int_{0}^{1}1\,dx\] is Riemann integrable.