Prove that if $f$ is integrable on $[a,b] \implies cf$ integrable on $[a,b]$ and $\int_{a}^{b} {cf}=c \int_{a}^{b} {f}$ 
Prove that if $f$ is Darboux-integrable on $[a,b] \implies cf$ Darboux-integrable on $[a,b]$ and $\int_{a}^{b} {cf}=c \int_{a}^{b} {f}$

My attempt: 
$\forall \epsilon >0$ $\exists P$ partition of $[a,b]: U(f,P)-L(f,P)< \epsilon$
Let $P=\{t_0,...,t_n\}, I_i=[t_{i-1},t_i], m_i=\inf_{I_i}{f}, M_i=\sup_{I_i}{f}$
$\implies L(f,P)=\sum_{i=1}^{n}{m_i(t_i-t_{i-1})},  U(f,P)=\sum_{i=1}^{n}{M_i(t_i-t_{i-1})}$
Obviously we have that $\inf_{I_i}{cf}=cm_i, \sup_{I_i}{cf}=cM_i$
$\implies L(cf,P)=c \sum_{i=1}^{n}{m_i(t_i-t_{i-1})}=cL(f,P)$
And $U(cf,P)=c \sum_{i=1}^{n}{M_i(t_i-t_{i-1})}=cU(f,P)$


*

*If $c \ge 0:$ Let $\epsilon >0 \implies 0 \le U(cf,P)-L(cf,P)=c(U(f,P)-L(f,P)) \le \epsilon$
We can make the $\epsilon >0 $ as small as we want  $\implies c \epsilon$ as well $\implies cf$ is Darboux-integrable on $[a,b]$
Also, we have that $L(cf,P)=cL(f,P) \le c \int_{a}^{b} {f}dx \le cU(f,P)=U(cf,P)$
$\implies \int_{a}^{b} {cf}=c \int_{a}^{b} {f}$

*If $c \le 0:$ Let $\epsilon >0 \implies c \epsilon \le c(U(f,P)-L(f,P)) \le 0$
We can make the $\epsilon >0 $ as small as we want  $\implies c \epsilon \to 0 \implies cf$ is Darboux-integrable on $[a,b]$
Also, we have that $L(cf,P)=cL(f,P) \ge c \int_{a}^{b} {f}dx \ge cU(f,P)=U(cf,P)$
$\implies \int_{a}^{b} {cf}=c \int_{a}^{b} {f}$
Is this correct? I'm not sure especially about the part where $c \le 0$
 A: Your proof for $c \geq 0$ looks right, but for $c \leq 0$, you can use the fact that
\begin{equation*}
\begin{split}
\sup \{ca : a \in A\} = c \inf \{a : a \in A\},\\
\inf \{ca : a \in A\} = c \sup \{a : a \in A\}.
\end{split}
\end{equation*}
This implies
\begin{equation*}
\begin{split}
U(cf, P) = cL(f, P),\\
L(cf, P) = cU(f, P).
\end{split}
\end{equation*}
Finally,
\begin{equation*}
\begin{split}
U(cf, P) - L(cf, P) &= cL(f, P) - c(U(f, P))\\
&= -c(U(f, P) - L(f, P))
< -c\epsilon
\end{split}
\end{equation*}
for all $\epsilon > 0$, remembering that $c \leq 0$.
A: Your proof is right, but it may not be formal enough, specifically when you talk about making $\epsilon$ "as small as we want", you can use this equivalent definition of integrability instead:
$\forall \epsilon > 0\ \exists\ P$ Partition of [a,b] such that
$|U(f,P)-L(f,P)| < |c|\epsilon'=\epsilon$ 
and show (using the sums' expansion) that:
$|cU(f,P)-cL(f,P)|=|c||U(f,P)-L(f,P)| <|c|\epsilon'=\epsilon$
Wich implies $\int_{a}^{b} {cf}=c \int_{a}^{b} {f}$
A: This proof will be carried out by using the Riemann definition of integral and it does not require supremum-infimum laws.
Proof.
If $c=0$, then the statement is trivial. We will consider the case $c≠0$. 
Choose an $\varepsilon>0$. Since $f\in \mathcal{R}[a,b]$, there exists $δ_\varepsilon>0$ such that for any tagged partition $\dot{\mathcal{P}}=\{([x_{i-1},x_i],t_i)\}_{i=1}^n$ of $[a,b]$ with $‖\dot{\mathcal{P}}‖<δ_\varepsilon$,
\begin{equation}
\left|S(f;\dot{\mathcal{P}})-\int_a^b f\right|<\frac{\varepsilon}{|c|}.
\end{equation}
Since for the partition $\dot{\mathcal{P}}$,
\begin{equation}
S(cf;\dot{\mathcal{P}})=\sum_{i=1}^n cf(t_i)(x_i-x_{i-1})=c\sum_{i=1}^n f(t_i)(x_i-x_{i-1})=S(f;\dot{\mathcal{P}}),
\end{equation}
we have
\begin{equation}
\left|S(cf;\dot{\mathcal{P}})-c\int_a^b f\right|=|c|\left|S(f;\dot{\mathcal{P}})-\int_a^b f\right|<\varepsilon.
\end{equation}
Since $\varepsilon>0$ is arbitrary, by the uniqueness of the Riemann integral, this proves that $cf\in \mathcal{R}[a,b]$ with
\begin{equation}
∫_a^b cf=c∫_a^b f. \blacksquare
\end{equation}
A: Chapter 13 of Spivak's Calculus (Theorem 6) also states this claim without proof :

If $f$ is integrable on $[a,b]$. then for any number $c$, the function $cf$ is integrable on $[a,b]$ and $\displaystyle \int_a^b cf = c \int_a^b f$

The following proof uses the conventions established in Spivak's book.

Suppose $f$ is integrable on $[a,b]$. Then we have that:

*

*$\displaystyle\sup_PL(f,P)=\int_a^bf=\inf_PU(f,P)$
Next, by definition, for some arbitrary partition $P$ of $[a,b]$, we have:


*$L(f,P)=\displaystyle \sum_{i=1}^nm_i\cdot(t_i-t_{i-1})$ and $U(f,P)=\displaystyle \sum_{i=1}^nM_i\cdot(t_i-t_{i-1})$
, where $m_i=\inf\{f(x):t_{i-1} \leq x \leq t_i\}$ and $M_i=\sup\{f(x):t_{i-1} \leq x \leq t_i\}$.
Therefore, for any interval $[t_{i-1},t_i]$ of this partition, there is an $m_i^f$ and $M_i^f$ such that for for all $x \in [t_{i-1},t_i]$:


*$m_i^f \leq f(x)$,  $\displaystyle\inf_x f(x)=m_i^f$ and $M_i^f\geq f(x),\displaystyle \sup_xf(x)=M_i^f$
Supposing that $c \geq 0$, from 3. we have that:


*$cm_i^f \leq cf(x)$ and $cM_i^f\geq cf(x)$
Straightforward contradictions arise if $\displaystyle\inf_x cf(x) \neq cm_i^f$ and $\displaystyle \sup_x cf(x) \neq M_i^f$, so we must have:


*$\displaystyle m_i^{cf}=\inf_x cf(x) = cm_i^f$ and $\displaystyle M_i^{cf}=\sup_x cf(x) = M_i^f$
Because $[c\cdot f](x)=cf(x)$, 4. and 5. allow us to conclude:


*$L(cf,P)=\displaystyle\sum_{i=1}^n cm_i^f\cdot(t_i-t_{i-1})=c\sum_{i=1}^n m_i^f\cdot(t_i-t_{i-1})=cL(f,P)$


*$U(cf,P)=\displaystyle\sum_{i=1}^n cM_i^f\cdot(t_i-t_{i-1})=c\sum_{i=1}^n M_i^f\cdot(t_i-t_{i-1})=cU(f,P)$
Similar styled contradiction arguments as used in 5. will then give us that:


*$\displaystyle \sup_P L(cf,P)=\sup_P c\cdot L(f,P)=c \cdot \sup_PL(f,P)=c\int_a^bf$


*$\displaystyle \inf_P U(cf,P)=\inf_P c\cdot U(f,P)=c \cdot \inf_P U(f,P)=c\int_a^bf$
Firstly, note that 8. and 9. are equal to other, which means that $cf$ is integrable on $[a,b]$ (by definition). By convention, we call this value $\displaystyle \int_a^b cf$. Secondly, the value of this integral is equal to $\displaystyle c \int_a^b$. Together, these two facts give us:


*$\displaystyle \sup_P L(cf,P)=\int_a^b cf =\inf_P U(cf,P)=c\int_a^bf$...or simply $\displaystyle \int_a^b cf=c\int_a^bf$

Starting again just after 3., suppose instead that $c \lt 0$. Then we have that:


*$cm_i^f \geq cf(x)$ and $cM_i^f\leq cf(x)$
Using the aforementioned proof by contradictions, we can conclude that:


*$\displaystyle M_i^{cf}= \sup_x cf(x)=cm_i^f$ and $\displaystyle m_i^{cf}= \inf_x cf(x)=cM_i^f$
With 11. and 12., we then have that:


*$\displaystyle L(cf,P)=\sum_{i=1}^n cM_i^f\cdot(t_i-t_{i-1})=c\sum_{i=1}^n M_i^f\cdot(t_i-t_{i-1})=cU(f,P)$


*$\displaystyle U(cf,P)=\sum_{i=1}^n c m_i^f\cdot(t_i-t_{i-1})=c\sum_{i=1}^n m_i^f\cdot(t_i-t_{i-1})=cL(f,P)$
Now, when we go to take the supremum and infimum (respectively) of 13. and 14., we need to be a little careful. Because $c \lt 0$: $\displaystyle \sup_P cU(f,P) \neq c \sup U(f,P)$. Similarly, $\displaystyle \inf_P cL(f,P) \neq c \inf L(f,P)$. Instead, we actually have:
15.$\displaystyle \sup_P c U(f,P)=c \inf_P U(f,P)$
16.$\displaystyle \inf_P c L(f,P)=c \sup_P L(f,P)$
To see why this is so, let us consider the first case of $\displaystyle \sup_P cU(f,P)=c \inf_PU(f,P)$. Because $f$ is bounded, we know that the set $\{U(f,P)| $P$ \text { is a partition of } [a,b]\}$ is bounded. This means that a greatest lower bound exists: call it $\beta$. For all $P$, we have that $\beta \leq U(f,P)$. Well, if $c \lt 0$, then we have that $c\beta \geq U(f,P)$. The same proof by contradiction that we have alluded to before will show that $c \beta$ is also the smallest such value that is $\geq U(f,P)$ for any $P$. Therefore, $\displaystyle \sup_P cU(f,P)=c \beta= c \inf_PU(f,P)$. With this strategy, we can conclude that:


*$\displaystyle \sup_P L(cf,f)=c \inf_P U(f,P)=c \int_a^b f$


*$\displaystyle \inf_P U(cf,f)=c \sup_P L(f,P)=c \int_a^b f$
From 17. and 18., we see that $cf$ must be integrable over $[a,b]$: by convention, we call this value $\displaystyle \int_a^b cf$. And, finally, that


*$\displaystyle \int_a^b cf = c\int_a^b f$
In conclusion, we have that:

if $f$ is integrable over $[a,b]$, then for any $c \in \mathbb R: \displaystyle \int_a^b cf = c\int_a^b f$.

