I have proved the following statement:
"Suppose $f, g:[a, b]\to\mathbb{R}$ are Riemann integrable. Prove that $f + g$ is Riemann integrable on $[a, b]$ and $\int_{a}^{b} (f+g)=\int_{a}^{b} f+\int_{a}^{b} g$."
but I am a bit unsure about how to handle suprema and infima in the part where I prove that $\int_{a}^{b} (f+g)=\int_{a}^{b} f+\int_{a}^{b} g$ so I would be grateful if someone could check my proof out, especially in that part: comments are welcome.
My proof:
Let $P$ be a partition $x_0,\dots, x_n$ of $[a,b]$. Then \begin{align}U(f+g,[a,b])\leq U(f+g,P,[a,b])&=\sum_{j=1}^{n}(x_j-x_{j-1})\sup_{[x_{j-1},x_j]}(f+g)\overset{*}{\leq}\sum_{j=1}^{n}(x_j-x_{j-1})(\sup_{[x_{j-1},x_j]}f+\sup_{[x_{j-1},x_j]}g)\\ &=\sum_{j=1}^{n}(x_j-x_{j-1})(\sup_{[x_{j-1},x_j]}f)+\sum_{j=1}^{n}(x_j-x_{j-1})\sup_{[x_{j-1},x_j]}g\\&=U(f,P,[a,b])+U(g,P,[a,b])\tag{1}\end{align} and \begin{align}L(f+g,[a,b])\geq L(f+g,P,[a,b])&=\sum_{j=1}^{n}(x_j-x_{j-1})\inf_{[x_{j-1},x_j]}(f+g)\overset{*}{\geq}\sum_{j=1}^{n}(x_j-x_{j-1})(\inf_{[x_{j-1},x_j]}f+\inf_{[x_{j-1},x_j]}g)\\&=\sum_{j=1}^{n}(x_j-x_{j-1})(\inf_{[x_{j-1},x_j]}f)+\sum_{j=1}^{n}(x_j-x_{j-1})\inf_{[x_{j-1},x_j]}g\\&=L(f,P,[a,b])+L(g,P,[a,b])\tag{2}\end{align}
(* if $f, g : A\to\mathbb{R}$ are bounded functions, then $\sup_A (f + g)\leq\sup_A f+\sup_A g$ and $\inf_A (f + g)\geq\inf_A f+\inf_A g$).
Now let $\varepsilon>0$. Since the upper Riemann integral is the infimum of the upper Riemann sums and the lower Riemann integral is the supremum of the lower Riemann sums there exist partitions $P_1, P_2, P_3$ and $P_4$ of $[a,b]$ such that
$U(f,[a,b])+\frac{\varepsilon}{2}>U(f,P_1,[a,b])$,
$U(g,[a,b])+\frac{\varepsilon}{2}>U(g,P_2,[a,b])$,
$L(f,P_3,[a,b])>L(f,[a,b])-\frac{\varepsilon}{2}$, and
$L(g,P_4,[a,b])>L(g,[a,b])-\frac{\varepsilon}{2}$
so if $P_5$ is the partition of $[a,b]$ obtained by merging the lists that define $P_1, P_2, P_3$ and $P_4$ we have (recalling the inequalities with Riemann sums): $U(f,[a,b])+\frac{\varepsilon}{2}+U(g,[a,b])+\frac{\varepsilon}{2}> U(f,P_5,[a,b])+U(g,P_5,[a,b])\tag{3}$ and $L(f,P_5,[a,b])+L(g,P_5,[a,b])>L(f,[a,b])+L(g,[a,b]-\varepsilon\tag{4}.$
Putting $(1)$ and $(3)$ together we have:
$U(f+g,[a,b])\leq U(f,P_5,[a,b])+U(g,P_5,[a,b])<U(f,[a,b])+U(g,[a,b])+\varepsilon.$
Since this holds for every $\varepsilon>0$ it follows that $U(f+g,[a,b])\leq U(f,[a,b])+U(g,[a,b])\tag{5}.$
Putting (2) and (4) together we have:
$L(f+g,[a,b])\geq L(f+g,P_5,[a,b])\geq L(f,P_5,[a,b])+L(g,P_5,[a,b])>L(f,[a,b])+L(g,[a,b])-\varepsilon.$
Since this holds for every $\varepsilon>0$ it follows that $L(f+g,[a,b])\geq L(f,[a,b])+L(g,[a,b])\tag{6}.$
$U(f+g,[a,b])\geq L(f+g,[a,b])$ together with $(5)$ and $(6)$ imply that $L(f,[a,b])+L(g,[a,b])\leq L(f+g,[a,b])\leq U(f+g,[a,b])\leq U(f,[a,b])+U(g,[a,b])$ from which, since $f$ and $g$ are integrable by hypothesis, follows that: $$U(f+g,[a,b])=L(f+g,[a,b])=\int_{a}^{b}(f+g)=\int_{a}^{b}f+\int_{a}^{b}g,$$ as desired. $\square$