Proving the more partition points the smaller the Darboux integral gets

Question

I'm trying to formally prove this:

Let $P=\{a=x_0<x_1<...<x_n=b\}$ be a a partition of $[a,b]$ which divides $[a,b]$ into $n$ equal sub intervals like so:

$x_i=a+\frac{i}{n}(b-a)\ \ \forall i\in\{0,1,...,n\}$

if:

$|U_{f,p}-L_{f,p}|<\epsilon\ \ $ (where $U_{f,p},L_{f,p}$ are Darboux upper\lower sums)

Then any partition $\tilde{P}$ of $[a,b]$ with $\lambda(\tilde{P})<\lambda(P)$ (where $\lambda(P)$ is the longest sub interval in $P$)

will also give:

$|U_{f,\tilde{p}}-L_{f,\tilde{p}}|<\epsilon\ \ $

This makes sense because any different partition must have more division points and that would make $|U_{f,p}-L_{f,p}|$ even smaller, but I'm having trouble showing it.

Can anyone give me any suggestions on this?

Answer 1

The lower sum for Darboux integral is defined as

$$ L_{f,P} = \sum_{x_i \in P} (x_{i+1} - x_i) \inf_{t \in [x_i, x_{i+1}]} f(t) $$

If we subdivide a single interval in a partition $P$, say $[x_0, x_1]$, into two subintervals $[x_0, x_{1/2}], [x_{1/2}, x_1]$, then we obviously have

$$ (x_1 - x_0)\inf_{t \in [x_0, x_1]} f(t) \leq (x_{1/2} - x_0)\inf_{t \in [x_0, x_{1/2}]} f(t) + (x_1 - x_{1/2}) \inf_{t \in [x_{1/2}, x_1]} f(t) $$ since the infimum over larger interval is necessarily no larger than infimum over subinterval.

Same works for upper Darboux sums, and since upper Darboux sum is obviously no smaller than lower Darboux sum, by taking a finer partition you can only get them closer to each other.

Answer 2

If $P\subset P'$ then $|U(P)-L(P)|\geqslant|U(P')-L(P')|$. But the fact that $\lambda(P)\geqslant\lambda(P')$ does not imply that $|U(P)-L(P)|\geqslant|U(P')-L(P')|$.

For a counterexample, consider the function $f$ such that $f(x)=0$ for $0\leqslant x\leqslant\frac12$, $f(x)=6x-3$ if $\frac12\leqslant x\leqslant\frac23$ and $f(x)=3-3x$ if $\frac23\leqslant x\leqslant1$, thus $f$ is the affine interpolation between the points $(0,0)$, $(\frac12,0)$, $(\frac23,1)$ and $(1,0)$. The regular partitions $P=(0,\frac12,1)$ and $P'=(0,\frac13,\frac23,1)$ are such that $\lambda(P)=\frac12\gt\frac13=\lambda(P')$ but $U(P)=\frac12$, $L(P)=L(P')=0$ and $U(P')=\frac23$ hence $|U(P)-L(P)|=\frac12\lt\frac23=|U(P')-L(P')|$.

Edit: If $f$ is monotone and continuous and $P_n$ is the regular partition of $[0,1]$ with $n+1\geqslant2$ points then $U(P_n)-L(P_n)=\frac1n|f(1)-f(0)|$. In particular, the sequence of general term $U(P_n)-L(P_n)$ is nonincreasing.