# A property of Darboux sum

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.

• 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. – RRL Mar 17 at 13:12
• 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. – RRL Mar 17 at 13:34
• I have not understood the way I should follow. I already obtained the last inequality using a refinement of $P_\epsilon$. – Nameless Mar 17 at 13:54
• – Paramanand Singh Mar 18 at 19:25
• Interesting, thank you @ParamanandSingh. – Nameless Mar 19 at 20:51

## 1 Answer

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

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

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