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I have a particular problem and its about Darboux sums and integrable functions. Here is a statement that is False, but want to demonstrate with example why its false. Here is the statement:

Let P be a partition, defined by, P ={1,1.3,2} be a partition of the closed interval [1,2]. Let f(x) = 0 for the range x<1.5, and let f(x) = 4 for x>= 1.5 . The the upper sum U(f,P) = 2.8 and the lower sum L(f,P) = 0. Therefore, function f is NOT integrable on [1,2] by the Darboux definition of the definite integral.

SO this statement is false. This function is integrable over this interval of [1,2].

My undestanding of the Darboux definition for integrable is that of, IF one can find a particular partition of a given interval such that L(f,P) = U(f,P), then the function is integrable over that interval by the Darboux definition.

I now want to find a partition of [1,2] such that, L(f,P) = U(f,P).

But I am having difficulty figuring out the partition, choosing the numbers for this to work.

Hope someone can figure it out.

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  • $\begingroup$ In general it will be Impossible to find a partition such that $L(f,P)=U(f,P)$, in particular this applies to any strictly monotonic function $f$. Note that integrability of $f$ in the sense of means that $\sup\{L(f,P) \vert P\in\mathcal{P}\}=\inf\{U(f,P) \vert P \in\mathcal{P}\}$, where $\mathcal{P}$ is the set of all partitions of the given interval. $\endgroup$ – Jens Schwaiger Apr 21 '18 at 6:04
  • $\begingroup$ Hi, if it will be that difficult to find "THE" partition of all possible partitions on this interval such that sup = inf, then how would one show that the function in this question actually is integrable by the Darboux definition? $\endgroup$ – Palu Apr 21 '18 at 15:09
  • $\begingroup$ Note that for partitions $P,Q$ with $P\subseteq Q$ the inequalities $L(P,f)\leq L(Q,f)\leq U(Q,f)\leq U(P,f)$. So one only needs to consider partitions $P$ wich contain $1.5$ as one point. But then $L(f,P)=0+\frac12 4=U(f,p)$. Thus for example $P=\{1,1.5,2\}$ is suitable. $\endgroup$ – Jens Schwaiger Apr 21 '18 at 19:02
  • $\begingroup$ Hi Jens Schwaiger, thanks for your reply. I think i get it. But when computing the U(f,P) is the upper sum for the sub-interval [1,1.5) is this equal to zero because we have an open interval at x=1.5. If that is the case we choose the left-end point which would give zero. Just hoping you will clarify if I got this idea correct. $\endgroup$ – Palu Apr 21 '18 at 21:25
  • $\begingroup$ When calculating the expressions $L(f,P)$ and $U(f,P)$ the suprema and infima in the subintervals are always taken with respectively to the closed subintervals. Compare for instance math.stackexchange.com/questions/2723771/…. $\endgroup$ – Jens Schwaiger Apr 22 '18 at 3:02

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