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- Apr 13, 2013

- 3,718

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??