Darboux and Riemann integrability: does the partition matter? When we say that a function is Riemann or Darboux integrable, do the partitions matter? In other words, suppose that we have a closed interval $[a,b]$ and a function $f$ that is bounded on said interval. If one can find only one partition such that $U(f)=L(f)$, would that be enough to show that the function $f$ is integrable on $[a,b]$? Or does the equality have to hold regardless of the choice of partition?
(Note: here $U(f)$ is the upper Darboux integral and $L(f)$ is the lower Darboux integral).
 A: The upper and lower Darboux integrals do not depend on the partitions, they are the infimum and supremum of all upper and lower Darboux sums for all possible partitions of your block. That is, $U(f)=\inf_{P}U(f,P)$ and $L(f)=\sup_{P}L(f,P)$. We know that for   any pair of partitions $L(f,P)\leqslant U(f,P')$. If for some partition, $L(f,P)=U(f,P)$, then it follows that $f$ is integrable, for it will follow $L(f)=\sup_P L(f,P)=U(f,P)\geqslant \inf_P U(f,P)=U(f)$. Since we always have $L(f)\leqslant U(f)$, we have $L(f)=U(f)$ and $f$ is Riemann integrable.
A: If you find a partition with that property, then you have proved integrability. Probably, though, you meant to say $U(f,P)=L(f,P)$, since the upper and lower Darboux integrals are not defined in terms of a single partition.
The point here is that $L(f,P)\le\underline\int f\le\overline\int f\le U(f,P)$ for any $P$, and to be Darboux integrable means that $\underline\int f=\overline\int f$.
A: If you've found single partition with identical upper and lower Darboux sums, then your function is piecewise constant, and those sums are exactly its integral.
However, most functions are of course not piecewise constant so this happens only in very special case.
If you're speaking of the upper and lower Darboux integrals, then your question makes no sense, because $U(f)$ and $L(f)$ do not depend on a particular choice of partition -- they are defined as the infimum / supremum over all possible partitions.
