I'm trying to show that:

$$\overline{I}:=\inf _P S(f;P)=\lim_{\lambda (P)\to 0}S(f;P)$$

where $P$ is a generic partition (made by $n$ points) of the interval $[a,b]$, $f$ is bounded on $[a,b]$, $S(f;P):=\sum _{i=1}^n\sup_{\Delta_i}f(x) \cdot \Delta x_i $ upper Darboux sum of $f$ on partition $P$ with mesh $\lambda (P)$.

Facts already known and proved:

a) $s(f;P_1):=\sum _{i=1}^{n_1}\inf_{\Delta_i}f(x) \cdot \Delta x_i \leq S(f;P_2)$, for all partitions $P_1$ and $P_2$.

b) $0\leq S(f;P)-S(f;\widetilde{P})\leq \omega (f;[a,b])\cdot (\Delta x_{k_1}+...+\Delta x_{k_m}) $, where $\widetilde{P} $ is a generic refinement of $P$, $ \omega (f;[a,b]):=\sup _{x',x'' \in [a,b]}|f(x')-f(x'')| $ and $\Delta x_{k_1},...,\Delta x_{k_m} $ are all the intervals of $P$ which contain points only in $\widetilde{P} $.

My attempt to demonstrate the statement:

$ \overline{I} $ is well defined thanks to a). Being an $\inf $, this implies that $\forall \epsilon >0$ there exists a partition $P_\epsilon $ such that:

$$\overline{I} \leq S(f;P_\epsilon) \leq \overline{I} + \epsilon$$

to conclude, it would be enough for me to show that any partition $ P $ with a mesh that is narrower than that of $ P_\epsilon $ leads to $S(f;P) \leq S(f;P_\epsilon) $, but all I managed to get (using also b)) was this:

$$S(f;P) \leq S(f;P_\epsilon) +\omega (f;[a,b])\cdot m\cdot \lambda (P)$$

where $m$ is the number of points in $P_{\epsilon} $ and $\lambda (P) $ is the mesh of the new partition $P$.

I can't do better. A little help, please?

Thanks in advance.

  • $\begingroup$ See here for a proof when the integral exists. The part of the argument for the upper sums converging to the upper Darboux integral is all you need for this question. $\endgroup$ – RRL Mar 17 at 13:12
  • $\begingroup$ You will see you are almost there. Take the partition $P$ with sufficiently small mesh and form a common refinement with $P_\epsilon$ to make the needed estimates. $\endgroup$ – RRL Mar 17 at 13:34
  • $\begingroup$ I have not understood the way I should follow. I already obtained the last inequality using a refinement of $P_\epsilon$. $\endgroup$ – Nameless Mar 17 at 13:54
  • 1
    $\begingroup$ See math.stackexchange.com/a/2047959/72031 $\endgroup$ – Paramanand Singh Mar 18 at 19:25
  • $\begingroup$ Interesting, thank you @ParamanandSingh. $\endgroup$ – Nameless Mar 19 at 20:51

You are on the right track. To help you finish, recall that it is to be shown for any $\epsilon > 0$ there exists a partition $P$ such that $|S(f;P) - \bar{I}| < \epsilon$ or equivalently

$$\tag{*}\bar{I} \leqslant S(f;P) \leqslant \bar{I} + \epsilon,$$

when $\lambda(P) < \delta$.

As you showed, now using $\epsilon/2$ instead of $\epsilon$, there exists a partition $P_\epsilon$ such that

$$\overline{I} \leqslant S(f;P_\epsilon) \leqslant \overline{I} + \frac{\epsilon}{2}$$

We want to introduce any partition $P$, not necessarily a refinement of $P_\epsilon$, and produce $\delta$ such that (*) is satisfied if $\lambda(P) < \delta$. Next, we construct a partition $\widetilde{P}$ that is a common refinement of $P$ and $P_\epsilon$ by adding the $m$ interior points of $P_\epsilon$ to $P$. Now, as you came close to reasoning, it follows that

$$S(f;P) \leqslant S(f;\widetilde{P}) + \omega (f;[a,b])\cdot m\cdot \lambda (P)$$

Further insight into why this is true can be obtained either by drawing a picture or following my argument here.

Since $\widetilde{P}$ is a refinement of $P_\epsilon$, we have $S(f;\widetilde{P}) \leqslant S(f;P_\epsilon)$, and it follows that

$$S(f;P) \leqslant S(f;P_\epsilon) + \omega (f;[a,b])\cdot m\cdot \lambda (P) \leqslant \overline{I} + \frac{\epsilon}{2}+ \omega (f;[a,b])\cdot m\cdot \lambda (P)$$

By choosing $\delta = \epsilon/ (2 \cdot m \cdot \omega (f;[a,b]))$, it follows that if $\lambda(P) < \delta$, then

$$\bar{I} \leqslant S(f;P) \leqslant \bar{I} + \epsilon$$

  • $\begingroup$ Thank you very much. (+1) $\endgroup$ – Nameless Mar 19 at 20:50
  • $\begingroup$ @Nameless: You're welcome. $\endgroup$ – RRL Mar 19 at 22:09

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