Let $c \in \Bbb R.$ Since each $L_k$ is measurable, if we let $A_k = \{ x \mid L_k(x) >c\}$ then these sets are measurable. Further since each set is bounded, they are finitely $\mu$-measurable (known result). If we define for any sets $X, Y,$
$$ d(X,Y) = m^*( (X\setminus Y) \cup (Y \setminus X)) $$
Where $m^*$ is the Lebesgue outer measure. Then if $A = \{ x \mid L(x) > c \}$ and $B = \{ x \mid f(x) > c \},$
$$ d(A_k,B) \leq d(A_k,A) + d(A,B) \rightarrow 0, $$
since $d(A,B)=0,$ by monotonicity of $m^*.$ Thus it follows that $B$ is measurable, by the definition provided in the book.
Denote the upper and lower Riemann integrals of $f$ by $\overline \int_I f$ and $\underline \int_I f$ respectively.
We have that $U(f+g,P) \le U(f,P) + U(g,P)$ and $L(f,P)+L(g,P)\le L(f+g,P)$. Thus, $$ L(f,P)+L(g,P)\le L(f+g,P)\le U(f+g,P) \le U(f,P)+U(g,P). $$
Note that the right-hand inequality gives us that, for each partition $P$, $$ \overline \int_I (f+g) \le U(f,P) + U(g,P), $$ so that $$ \overline \int_I (f+g) \le \overline \int_I f + \overline \int_I g .$$ Similarly, $$ \underline\int_I f + \underline\int_I g \le \underline\int_I (f+g). $$
We know that $f$ and $g$ are Riemann-integrable, which means that their upper and lower Riemann integrals are the same. This allows us to conclude that $$ \int_I f + \int_I g = \overline\int_I (f+g) = \underline\int_I (f+g),$$ which is what we needed to show.
Best Answer
I can provide a brief outline and you can develop that into a proper proof. First you need to understand that the definition of Darboux integral as given in your post is equivalent to the formulation given in most textbooks. Thus given a bounded function $f:[a, b] \to\mathbb {R} $ we take an arbitrary partition $$P=\{a=x_0,x_1,x_2,\dots,x_n=b\}, x_{i-1}<x_{i},i=1,2,\dots,n$$ of $[a, b] $ and define Darboux sums $$L(f, P) =\sum_{i=1}^{n}m_{i}(x_i-x_{i-1}),\, U(f,P)=\sum_{i=1}^{n}M_{i}(x_i-x_{i-1})$$ where $$M_i=\sup_{x\in[x_{i-1},x_i]}f(x),\, m_i=\inf_{x\in[x_{i-1},x_i]}f(x)$$ Let $\mathcal{P}[a, b] $ denote the set of all possible partitions of interval $[a, b] $. The Darboux sums are themselves bounded ($L(f, P) \leq M(b-a), U(f, P) \geq m(b-a) $ where $M, m$ are supremum and infimum of $f$ on $[a, b] $) and thus $$\overline{J} =\inf_{P\in\mathcal{P} [a, b]} U(f, P), \, \underline{J} =\sup_{P\in\mathcal{P} [a, b]} L(f, P) $$ exist. You need to show that $$\underline{J} =\sup_{g\leq f, \text{ piecewise constant}}\text{p.c.}\int_{a}^{b}g(x)\,dx,\,\overline{J}=\inf_{h\geq f, \text{ piecewise constant}} \text{p.c.}\int_{a}^{b}h(x)\,dx$$ which establishes the equivalence of definitions of Darboux integral as given by Tao and as given in other textbooks.
Once this is done you need to show that $$\underline{J} =\lim_{|P|\to 0}L(f,P),\,\overline{J}=\lim_{|P|\to 0}U(f,P)$$ This is already done in this answer.
Since a Riemann sum is always sandwiched between upper and lower Darboux sums, it follows that if $f$ is Darboux integrable then both upper and lower Darboux sums tend to a common limit as $|P|\to 0$ and therefore so do the Riemann sums and the value of the integral defined by these approaches is also same.
To go the reverse way (from Riemann to Darboux) one just needs to prove that one can find a Riemann sum as near to a Darboux sum as we please by choosing the tag points $t_i\in[x_{i-1},x_i]$ such that $f(t_i) $ is near $M_i$ (or $m_i$ as needed). And thus if Riemann sums tend to a given value then the Darboux sums also do the same.