Approximating Darboux integrals with continuous functions Let $f:[a,b]\rightarrow\mathbb{R}$ be a bounded function.  Show that the upper Darboux integral of $f$ is equal to the infimum of the Riemann integral of $g$ over all continuous functions $g\geq f$.
I'm not sure how to approximate an arbitrary bounded function with a continuous function.  One thought I had is that that upper Darboux integral is itself an infimum, namely the infimum of upper Darboux sums.  So we are basically trying to show that the infimum of two sets are equal to each other.  But I'm not sure what the conditions are under which two sets have the same infimum, though my question here is relevant to that.
 A: Given $\epsilon > 0$ there exists a partition $P = (x_0,x_1,\ldots,x_n)$ and upper Darboux sum $U(P,f)$ such that 
$$\overline{\int_a}^b f \leqslant U(P,f) < \overline{\int_a}^b f + \epsilon/2$$
Note that the upper Darboux sum is the integral of a step function $\phi_P = \sum_{j=1}^NM_j \mathbb{1}_{[x_{j-1},x_j)}$ with $M_j = \sup_{x \in [x_{j-1},x_j]} f(x)$, that is
$$U(P,f) = \int_a^b\phi_P = \sum_{j=1}^NM_j (x_j- x_{j-1})$$
For each point $x_j$ we can take a small interval $[x_j- \delta,x_j]$ and straight line joining $(x_j-\delta,\phi_P(x_j- \delta))$ and $(x_j,\phi_P(x_j))$, thereby constructing a continuous, piecewise linear function $g$ such that $f \leqslant \phi_P \leqslant g$ and $0 \leqslant \int_a^b (g - \phi_P) < \epsilon/2$.  We then have
$$\overline{\int_a}^b f \leqslant \int_a^b \phi_P = \int_a^b g - \int_a^b (g-\phi_P) < \overline{\int}_a^b f + \epsilon/2$$
Thus, for any $\epsilon > 0$ there is a continuous function $g \geqslant f$ such that
$$\tag{1} \overline{\int_a}^b f \leqslant \int_a^bg < \overline{\int_a}^b f +\epsilon$$
It is easy to show that for $\mathcal{G} = \{g:[a,b] \to \mathbb{R} \, | \, g \geqslant f, \, g \text{ continuous}  \}$, we have
$$\tag{2} \overline{\int_a}^b f \leqslant \inf_{g \in \mathcal{G}} \int_a^b g,$$
since upper sums of $g$ dominate upper sums of $f$.
A strict inequality in (2) would contradict (1) which shows there exist continuous functions $g \in \mathcal{G}$ such that $\int_a^b g$ is arbitrarily close to $\overline{\int_a}^b f$ (as $\epsilon \to 0$).
Thus,
$$\overline{\int_a}^b f = \inf_{g \in \mathcal{G}} \int_a^b g$$
