If Darboux (Riemann equivalent), then Lebesgue? Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be compact, define $$D^+(f):= \inf\left\{\int t:t\geq f, t= \text{step function}\right\}$$ $$D^-(f):= \sup\left\{\int t:t\leq f, t= \text{step function}\right\}$$ If $D_+(f)=D_-(f),$ we call $f$ Darboux integrable (equivalent to Riemann integrable) and the common value the Darboux integral of $f$.
How would I show that if $f$ is Darboux integrable, then $f$ is Lebesgue integrable, and $D^+(f)=D^-(f)=\int f$
 A: For a bounded interval consider WLOG   $f :[a,b] \to \mathbb{R},$ a nonnegative Riemann integrable function. Then
$$\int_a^b f(x) \, dx  = \sup_{\phi  \leqslant f} \int_a^b\phi(x) \, dx = \inf_{\psi  \geqslant f} \int_a^b\psi(x) \, dx ,$$
where $\phi$ and $\psi$ are step functions.  This is straightforward to show using upper and lower Darboux sums and integrals.
Since any step function is a simple function, we have for simple functions $\hat{\phi}$ and $\hat{\psi}$
$$D^-(f) = \sup_{\phi  \leqslant f} \int_a^b\phi(x) \, dx \leqslant \sup_{\hat{\phi}  \leqslant f} \int_{[a,b]}\hat{\phi} \leqslant \inf_{\hat{\psi}  \geqslant f} \int_{[a,b]}\hat{\psi}  \leqslant \inf_{\psi  \geqslant f} \int_a^b\psi(x) \, dx = D^+(f),$$
and Riemann integrability, $D^-(f) = D^+(f),$ implies Lebesgue integrability with
$$\sup_{\hat{\phi}  \leqslant f} \int_{[a,b]}\hat{\phi}  = \int_{[a,b]}f= \int_a^bf(x) \, dx .$$
The converse is not true.  Lebesgue integrability on a bounded interval does not imply Riemann integrability. The Dirichlet function is a counterexample.
