# Show that $L(f)=\sup\{L(P,f):P \in \mathcal P_{c}\}$,where $c\in \left(a,b\right)$

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.

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 $$L(P',f)$$.

• Well $L(P,f) \le L(P',f)$ – masaheb Jan 16 at 16:19
• @masaheb: correct. do you know how to go on? – user9464 Jan 16 at 16:47
• $P'$ is a refinement of $P$ and so $L(P,f) \le L(P',f)$,that's all I know. – masaheb Jan 16 at 16:48
• @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? – user9464 Jan 16 at 16:51
• Then I guess we need to use the second part – masaheb Jan 16 at 16:55