"Show that if $f$ is integrable on $[a,b]$, then $cf$ is integrable on $[a,b]$, where $c\in \Bbb{R}$"
I think we need to look at $c\geq 0$, and $c\leq 0$ separately. Let $c\geq 0$ and $P$ any partition of $[a,b]$, and let $m_i$ and $m_i'$ be the infima of $f$ and $cf$ on some sub-interval $\Delta x_i$ of $P$. Now $$m_i\leq f \implies cm_i\leq cf$$ $$m_i' \leq cf$$
which means that $cm_i \leq m_i'$ ( where $i=1,…,n$ ). Multiplying by $\Delta x_i$ gives $$c \cdot \Delta x_i m_i\leq\Delta x_i m'_i \implies c\cdot L(P,f)\leq L(P, cf).$$
Similarly, let $M_i$ and $M'_i$ be the suprema of $f$ and $cf$, again on $\Delta x_i$. This now yields $$cM_i \geq M'_i\implies c \cdot \Delta x_iM_i\geq M'_i \Delta x_i \implies c\cdot U(P,f) \geq U(p,cf).$$
Which gives the following inequality $$c\cdot L(P,f) \leq L(P,cf) \leq U(P,cf) \leq c\cdot U(p,f). ( 1)$$
If this is correct, I see this as similar to the proof of the integrability of the sum of two integrable functions. However, I'm not sure what the next step here should be. I think I should maybe use $$c \cdot U(P,f) – c \cdot L(P,f) < c \cdot \epsilon.$$
along with $(1)$. Any hints are appreciated.
Best Answer
Firstly let $\epsilon\gt 0$ be given. Then there exists a partition $P$ of $[a,b]$ such that $U(p,f)-L(P,f)\lt\frac{\epsilon}{c}$
Let $c\gt 0\\$ and $\ P$ be that partition of [a,b], and let $m_i\\$ and $m_i^{'}\\$ be the infima of f and cf on some sub-interval $Δx_i\\$ of P.
Now, $\\inf (cf)=c\cdot inf(f)\implies c \cdot \Delta x_iM_i= M'_i \Delta x_i \implies c\cdot L(P,f)= L(P,cf) \\ $
Similarly, let $\ M_i\\$ and $\ M_i^{'}\\$ be the suprema of f and cf, again on $\ Δx_i\\$. This now yields $\\cM_i = M'_i\implies c \cdot \Delta x_iM_i= M'_i \Delta x_i \implies c\cdot U(P,f) = U(p,cf).$
Therefore $\\ U(p,cf)-L(P,cf)= c[ U(p,f)-L(P,f)] \lt c\cdot \frac{\epsilon}{c}=\epsilon \\$ for a partition $P$.
Thus being a sufficient condition for integrability, $cf$ is integrable.
When $c\lt0$ it can be proven similarly. And for $c=0$ $cf(x)=0$ and is integrable.