Prof that the lower Darboux integral is monotonic

Question

Suppose the sequence of partitions $P_n={x_0,...,x_n}$ is given by $x_i=i/n$, and $f$ is continuous on $[0,1]$. I've worked through enough examples to suspect that $L(P_n,f)$, the lower Darboux integral, is monotonic in $n$. I now need to prove it, and I can see that the infs on each interval will only grow larger as we choose smaller intervals. But finding a rigorous connection from that to the claim is proving challenging. I thought about the fact that, for constant $a$, we know that $\sum_{i=1}^n \frac{a}{n}=\sum_{i=1}^{n+1} \frac{a}{n+1}$ but can't see how I could extend that fact to this more general setting.

Answer 1

I actually think I just found a counterexample: $f$ is 0 on $(0,1/2)$ and is 1 on $(1/2, 1)$. Then $L(P_2,f)=1/2$ and $L(P_3,f)=1/3$, but thereafter the lower Darboux integral converges to 1/2.

Answer 2

The lower Darboux sums will increase if the partitions in the sequence are successive refinements. That is, if $P_m \subset P_{m+1}$ then $L(P_{m+1},f) \geqslant L(P_m,f)$.

An example with uniform spacing where lower sums are monotone is the sequence of dyadic partitions:

$$P_n = \left(0, \frac{1}{2^n}, \frac{2}{2^n}, \ldots, 1 \right),$$

since every subinterval of $I'$ of $P_{n+1}$ is contained in some subinterval of $I$ of $P_n$ and if $I' \subset I$ then $\inf_{I'} f(x) \geqslant \inf_I f(x)$.

This monotone behavior of lower sums is not always true for partitions of the form $(0,1/n, 2/n, \ldots,1)$ and it is relatively easy to produce a counterexample where $f$ is continuous:

$$f(x) = \begin{cases} 0 , \,\,\,\ 0 \leqslant x \leqslant 1/3 \\ 6(x - 1/3), \,\,\, 1/3 < x < 1/2\\1, \,\,\,\,\, 1/2 \leqslant x \leqslant 1\end{cases}$$