Upper-Lower sum of Riemann Integral

[evinda]

Hello! (Wave)

I am looking at the proof that if $f$ is integrable and $k \in \mathbb{R}$,then $kf$ is also integrable and $\int_a^b{(kf)}=k \int_a^b{f}$.

The following identity is used at my textbook:
$$U(kf,P)=\left\{\begin{matrix}
k \cdot U(f,P), \text{ if } k>0\\
k \cdot L(f,P), \text{ if } k<0
\end{matrix}\right.\text{ and } L(kf,P)=\left\{\begin{matrix}
k \cdot L(f,P), \text{ if } k>0\\
k \cdot U(f,P), \text{ if } k<0
\end{matrix}\right.$$

For $k>0$ it is like that: $U(kf,P)=\Sigma_{i=0}^{n-1}(t_{i+1}-t_i)sup(kf)([t_i,t_{i+1}])=k \cdot \Sigma_{i=0}^{n-1}(t_{i+1}-t_i)sup(f)([t_i,t_{i+1}])=k \cdot U(f,P)$For $k<0$,let $k=-m,m>0$.We have: $U(kf,P)=\Sigma_{i=0}^{n-1}(t_{i+1}-t_i)sup(m(-f))([t_i,t_{i+1}])=m \cdot \Sigma_{i=0}^{n-1}(t_{i+1}-t_i)sup((-f))([t_i,t_{i+1}])$

Is the last relation equal to $m \cdot \Sigma_{i=0}^{n-1}(t_{i+1}-t_i)inf(f)([t_i,t_{i+1}])$?? But,if it was like that,$U(kf,P)=m \cdot L(f,P)=-k \cdot L(f,P)$..Or am I wrong??

 

[ThePerfectHacker]

$$\sup_{[t_{k-1},t_k]}(-f) = - \inf_{[t_{k-1},t_k]}f$$

 

[evinda]

$$\sup_{[t_{k-1},t_k]}(-f) = - \inf_{[t_{k-1},t_k]}f$$

A ok.. :) And what's with $\inf_{[t_{k-1},t_k]}(-f)$ ? Is it equal to $- \sup_{[t_{k-1},t_k]}f$ ??

 

[ThePerfectHacker]

A ok.. :) And what's with $\inf_{[t_{k-1},t_k]}(-f)$ ? Is it equal to $- \sup_{[t_{k-1},t_k]}f$ ??

Yes. To see why here is an exercise.

Exercise: Let $A$ be a non-empty bounded set of real numbers. Define $-A = \{ -a ~ | a\in A\}$, the set of negatives of $A$. Show that $-A$ is a bounded set also, and $\sup(-A) = -\inf A$ and $\inf(-A) = -\sup A$.

 

[blue_raver22]

Hello! I can confirm that your approach is correct. The last relation should indeed be equal to $m \cdot \Sigma_{i=0}^{n-1}(t_{i+1}-t_i)inf(f)([t_i,t_{i+1}])$. This means that $U(kf,P)=-k \cdot L(f,P)$, which is in line with the identity given in your textbook. Keep up the good work!