Show that $\int_a^b {f+g}=\int_a^b f +\int_a^b g$
Proof:
Let $\epsilon \gt 0$ be given. There exist partitions $P,Q$ of $[a,b]$ such that $S_Pf-s_Pf \lt \frac \epsilon 2$ and $S_Qg-s_Qg \lt \frac \epsilon 2$ , by integrability of $f$ and $g$ on $[a,b] \qquad \qquad\mathbf *$
Now, $P\cup Q$ is a partition of $[a,b]$ and it's finer than $P$ and $Q$.
So,$$S_{P\cup Q}f-s_{P\cup Q}f \leq S_Pf-s_Pf\lt \frac \epsilon 2$$ $$S_{P\cup Q}g-s_{P\cup Q}g \leq S_Qg-s_Qg\lt \frac \epsilon 2$$
And, $$S_{P\cup Q}(f+g)\leq S_{P\cup Q}f +S_{P\cup Q}g$$ $$s_{P\cup Q}(f+g) \geq s_{P\cup Q}f +s_{P\cup Q}g$$ Hence, we get
$$S_{P\cup Q}(f+g)-s_{P\cup Q}(f+g)\leq S_{P\cup Q}f+ S_{P\cup Q}g- s_{P\cup Q}f-s_{P\cup Q}g\lt \epsilon$$
So, we proved $f+g$ is integrable on $[a,b].$
And the rest of the proof is remained for the reader. My attempt is:
We proved $S_{P\cup Q}(f+g)-s_{P\cup Q}(f+g)\lt \epsilon $
So, $$S_{P\cup Q}(f+g)-\epsilon \lt\int_a^b (f+g)\lt s_{P\cup Q}(f+g) +\epsilon$$
Also,
$$S_Pf- \frac \epsilon 2\lt \int_a^b f\lt s_Pf+\frac \epsilon 2$$
$$S_Qg- \frac \epsilon 2\lt \int_a^b g\lt s_Qg+\frac \epsilon 2$$
$$\therefore S_Pf+S_Qg-\epsilon\lt \int_a^bf+ \int_a^b g\lt s_Pf+s_Qg+\epsilon$$
So,
$$|\int_a^b(f+g)-\big (\int_a^b f+ \int_a^b g\big ) |\lt s_{P\cup Q}(f+g)-S_Pf-S_Qg+2\epsilon$$
Now how can I show that $s_{P\cup Q}(f+g)-S_Pf-S_Qg$ is so small to prove $\int_a^b {f+g}=\int_a^b f +\int_a^b g$ ?
By the way the proof starts with assuming $f$ and $g$ are integrable on$[a,b]$ see $:\quad (\mathbf *)$ Yes, the author assumes the integrability of $f$ and $g$ on $[a,b]$ and proves the integrability of $f+g$ on $[a,b]$. Is it the way this proof has to be done? Could I do it the other way around, that is,assuming the integrability of $f+g$ on $[a,b]$ and the proving the other one?
The integrability of $f+g$ says nothing about $f$ and $g$: For example, $f=-\chi_{{\bf{Q}}\cap[0,1]}$ and $g=\chi_{{\bf{Q}}\cap[0,1]}$.
For the first question, so for any point $t_{i}\in[x_{i-1},x_{i}]$, $i=1,...,n$, where $x_{0},...,x_{n}$ are partition points of $P\cup Q$, then \begin{align*} &\left|\sum_{i=1}^{n}(f+g)(t_{i})(x_{i}-x_{i-1})-\int_{a}^{b}f(x)dx-\int_{a}^{b}g(x)dx\right|\\ &\leq\left|\sum_{i=1}^{n}f(t_{i})(x_{i}-x_{i-1})-\int_{a}^{b}f(x)dx\right|+\left|\sum_{i=1}^{n}g(t_{i}(x_{i}-x_{i-1})-\int_{a}^{b}g(x)dx\right|\\ &<\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}\\ &=\epsilon, \end{align*} this proves that $\displaystyle\int_{a}^{b}(f+g)(x)dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx$.