Riemann integral

For a bounded real function on a closed finite interval, the Riemann integral exists when the supremum of its lower Darboux sums equals the infimum of its upper Darboux sums. Their common value is ${\int }_a^{b}f(x)dx$. Equivalently, for every $ϵ>0$ there is a partition of an interval whose upper-minus-lower sum is less than $ϵ$.^[ref]

For $f(x)=x$ on $[0,1]$, uniform partitions into $n$ pieces give lower and upper sums $1/2-1/(2n)$ and $1/2+1/(2n)$, proving that the integral is $1/2$. The Dirichlet function has lower sums zero and upper sums one on this interval, so it is not Riemann integrable. The definition explains why boundedness alone is insufficient.