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Assume $\mathcal P_{[a,b]}$ is the set of all partitions of $[a,b]$ and $\mathcal P_{c}$ is the set of all partitions of $[a,b]$ containing $c$,where $c\in \left(a,b\right)$,if $f:I \to \mathbb R$ is a bounded function over $I$,show that $$L(f)=\sup\{L(P,f):P \in \mathcal P_{c}\}$$


I know that

$$-M(b-a) \le L(P,f) \le L(f) \le U(f) \le U(P,f)\le M(b-a)$$

Where $M$ is a real number such that $\forall x \in [a,b]:-M \le f(x) \le M$

And $$ U\left(P_{},f\right)=\sum_{i=0}^{n-1}M_{i}\Delta x_i$$

$$ L\left(P_{},f\right)=\sum_{i=0}^{n-1}m_{i}\Delta x_i$$

With $M_{i}=\sup \{f(x):x \in [x_{i},x_{i+1}]\}$ and $m_{i}=\inf \{f(x):x \in [x_{i},x_{i+1}]\}$,$$U(f)=\inf \{ U\left(P,f\right):P \in \mathcal P_{[a,b]} \}$$ $$L(f)=\sup \{ L\left(P,f\right):P \in \mathcal P_{[a,b]} \}$$

So I used the fact that $f$ is bounded on $I$,but I don't know how to use $c$ to conclude the desired answer.

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The conclusion has nothing to do with the upper sum.

Hint.

Prove two inequalities: $$L(f)=\sup\{L(P,f):P \in \mathcal P_{[a,b]}\}\le\sup\{L(P,f):P \in \mathcal P_{c}\}\tag{1} $$ $$L(f)=\sup\{L(P,f):P \in \mathcal P_{[a,b]}\}\ge\sup\{L(P,f):P \in \mathcal P_{c}\}\tag{2} $$

Note that (2) is trivial since $\mathcal{P}_c\subset \mathcal{P}_{[a,b]}$.

To show (1), note that for any partition $P$ of $[a,b]$ that is not in $\mathcal{P}_c$, you can always add a point $c$ to get a new partition $P'\in\mathcal{P}_c$. Now compare $L(P,f)$ and $L(P',f)$.

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  • $\begingroup$ Well $L(P,f) \le L(P',f)$ $\endgroup$ – masaheb Jan 16 at 16:19
  • $\begingroup$ @masaheb: correct. do you know how to go on? $\endgroup$ – user9464 Jan 16 at 16:47
  • $\begingroup$ $P'$ is a refinement of $P$ and so $L(P,f) \le L(P',f) $,that's all I know. $\endgroup$ – masaheb Jan 16 at 16:48
  • $\begingroup$ @masaheb: good. Then you have $L(P,f)\le L(P',f)\le\textrm{the right hand side of (1)}$. Can you see that how (1) follows? $\endgroup$ – user9464 Jan 16 at 16:51
  • $\begingroup$ Then I guess we need to use the second part $\endgroup$ – masaheb Jan 16 at 16:55

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