If $f$ and $g$ are Riemann integrable in $[a,b]$ then $f+g$ is Riemann integrable in $[a,b]$ and $$\int_a^b\left[f(x)+g(x)\right]dx=\int_a^b f(x)\,dx+\int_a^b g(x)\,dx.\\ $$
Also, if $\lambda\in\mathbb{R}$, $\lambda f$ is Riemann integrable and
$$\lambda\int_a^bf(x)\,dx=\int_a^b\lambda f(x)\,dx.$$
Proof. Fix $\epsilon>0$ and let $u_1,v_1$ and $u_2,v_2$ be step functions of compact support such that
$$\begin{aligned}
&u_1\leq f\leq v_1, \qquad \int_a^b\left(v_1-u_1\right)\leq\epsilon/2.\\
&u_2\leq g\leq v_2, \qquad \int_a^b\left(v_2-u_2\right)\leq\epsilon/2.
\end{aligned}$$
Then
$$u_1 + u_2\leq f+g\leq v_1+v_2,\qquad
\begin{aligned}
&\int_a^b\left(\left(v_1+v_2\right)-\left(u_1+u_2\right)\right)=\\&\int_a^b\left(v_1-u_1\right)+\int_a^b\left(v_2-u_2\right)\leq\epsilon.
\end{aligned}$$
which shows the integrability of $u_1+u_2$; Also if $\alpha=\int_a^b f$ and $\beta=\int_a^b g$, we have
$$\int_a^b u_1\leq\alpha\leq\int_a^bv_1,\qquad \int_a^b u_2\leq\beta\leq\int_a^bv_2$$
and also
$$\int_a^b u_1+\int_a^b u_2=\int_a^b\left(u_1+u_2\right)\leq\int_a^b(f+g)\leq\int_a^b\left(v_1+v_2\right)=\int_a^bv_1 +\int_a^bv_2.$$
these inequalities easily lead to
$$\color{red}{(?)}\,\,\,\left|\int_a^b\left(f+g\right)-\left(\alpha+\beta\right)\right|\leq\int_a^b\left(\left(v_1+v_2\right)-\left(u_1+u_2\right)\right)\leq\epsilon.$$
Since $\epsilon$ can be chosen arbitrarily, then $\\\int_a^b(f+g)=\alpha+\beta$.
$$\\ \\$$
For the second part of the theorem, the case where $\lambda=0$ is trivial. Fix $\epsilon>0$ and let $u,v\in S_c$ with $u\leq v$ and $\int_a^b\left(v-u\right)\leq\epsilon$; if $\lambda>0$ we have $\lambda u\leq \lambda f\leq \lambda v$, and $\int_a^b\left(\lambda v-\lambda u\right)\leq \lambda\epsilon$, if $\lambda<0$, we have $\lambda v\leq \lambda u$, and $\int_a^b\left(\lambda u-\lambda v\right)=-\lambda\int_a^b\left(v-u\right)\leq\left(-\lambda\right)\epsilon=|\lambda|\epsilon$; In either case $\lambda f$ is enclosed by two functions of $S_c$ with integrals that differ by less than $\,|\lambda|\epsilon$; Also, if $\alpha=\int_a^bf(x)\,dx$, we directly have
$$\color{red}{(?)}\,\,\,\left|\int_a^b\lambda f(x)\,dx-\lambda\alpha\right|\leq |\lambda|\epsilon,$$ since we can choose $\epsilon$ arbitrarly then $\int_a^b\lambda f(x)\,dx=\lambda\int_a^b f(x)\,dx$ holds.
$$\tag*{$\square$}$$
I am having some trouble understanding the proof of the second part of the above theorem. More specifically, how the author was able to obtain $\color{red}{(?)}$. Also, shouldn't it be $\left(-\lambda\right)\epsilon=\color{red}{-}|\lambda|\epsilon$ instead of $\left(-\lambda\right)\epsilon=|\lambda|\epsilon$? Thank you.
Best Answer
In the case the author is discussing, $\lambda <0$ so $|\lambda| = -\lambda$ and hence $$\left(-\lambda\right)\epsilon=|\lambda|\epsilon$$ as he writes.
As for the integral inequality (?), in both cases, whether $\lambda < 0$ or $\lambda > 0$, his logic applies directly, as he was able to choose the bounding functions $u$ and $v$ as described, and when $\lambda =0$ then equality holds since both sides are $0$.
Feel free to ask more detailed questions in the comments to this answer for further guidance.