How to prove that a function is Riemann integrable if and only if it is Darboux integrable? (Pg. 16 Exercise 1.1.22. in Terry Tao’s Measure Theory) Show that if $f:[a,b]\to\mathbb{R}$ is a bounded function, then it is Riemann integrable if and only if it is Darboux integrable such that its Darboux and Riemann integrals are equivalent. In other words
$$\int_a^b f(x) \ dx=\lim_{\|P\|\to 0}\mathcal{R}(f,\mathcal{P})\Longleftrightarrow\sup_{g\le f;\kern 0.1em\mathrm{piecewise \kern 0.1em constant}}\textrm{p.c.}\int_a^b g(x) \ dx =\inf_{h\ge f,\kern 0.1em\mathrm{piecewise \kern 0.1em constant}}\textrm{p.c.}\int_a^b h(x) \ dx.$$
Could anyone give me any hints?
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 $$P=\{a=x_0,x_1,x_2,\dots,x_n=b\}, x_{i-1}&lt;x_{i},i=1,2,\dots,n$$ 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.
The approximation for a Riemann integral is bounded above and below respectively by the Darboux sup and inf.  Therefore Darboux implies Riemann.

Riemann sums can be constructed as Darboux upper and lower sums.  Therefore Riemann implies Darboux.