# Counter-Example for Darboux Sums: “Finer” Partition with Greater Difference.

Let $$P = \{p_i\}_{i= 1, n}$$, $$P' = \{p'_j\}_{j = 1, m}$$ be partitions of an interval with max$$|p'_j| \le$$min$$|p_i|$$, i.e. all the sub-intervals of $$P'$$ are at least as short as all the sub-intervals of $$P$$.
Let $$f$$ be a bounded function on the interval.
Let $$U_{P, f}, L_{P, f}$$ be the upper and lower Darboux sums of $$f$$ on $$P$$ (and correspondingly on $$P'$$).

There is a proof here https://math.stackexchange.com/q/353810 in the lemma that $$U_{P', f} - L_{P', f} \le 3(U_{P, f} - L_{P, f} )$$.

I assume from this that it is not generally the case that $$U_{P', f} - L_{P', f} \le U_{P, f} - L_{P, f}$$ and I'm looking for a counter-example to show this is so. I do know that this tighter inequality holds when $$P'$$ is a refinement of $$P$$ but even so I cannot construct a counter-example.

After further thought .....

Let $$P = \{0, 1/2 , 1\}$$ and $$P' = \{0, 1/3, 2/3, 1\}$$ be partitions of $$[0, 1]$$. Define $$f$$ on $$[0, 1]$$ by $$f(7/12) = 1; f(1) = 1; f(x) = 0$$ otherwise.

Then $$U_{P', f} - L_{P', f} = 0 + 1/3 + 1/3 = 2/3 > 1/ 2 = U_{P, f} - L_{P, f}$$

• If you have answered your own question, shouldn't you post the answer as an answer, instead of appending it to the question? – bof Feb 20 at 9:43
• @bof I have once or twice in the past, but now wonder if it doesn't just look like going for the points ? – Tom Collinge Feb 20 at 10:57