If $f \le g$ on $[a,b]$ and both are bounded, then the upper integral of $f$ is less than that of $g$ How to prove the following:

If $f \le g, \forall x \in [a,b]$, f and g are both bounded on [a,b], then upper integral of f is less then or equal to upper integral of g. 

It is easy to show this when f and g have the same partition. But when they have the different partition, I think maybe we can use a new partition, which can refine both of them, but I do not know how to set up the inequality. 
 A: The upper integral of $h$ is
$$
                 \overline{\int_{a}^{b}}h(x)dx = \inf_{\mathcal{P}}\overline{S}_{\mathcal{P}}(h),
$$
where $\overline{S}_{\mathcal{P}}(h)$ is the supremum of all Riemann sums over a fixed partition $\mathcal{P}$ obtained by allowing the evaluation points to vary. For the same evaluation points, $S_{\mathcal{P}}(f)\le S_{\mathcal{P}}(g)$. Therefore,
$$
  S_{\mathcal{P}}(f)
            \le S_{\mathcal{P}}(g) 
            \le \overline{S}_{\mathcal{P}}(g).
$$
The far right side is independent of the evaluation points of $\mathcal{P}$. Therefore, $S_{P}(f)\le \overline{S}_{\mathcal{P}}(g)$, and the right side does not depend on evaluation points of $\mathcal{P}$. Taking the supremum over all evaluation points then gives $\overline{S}_{\mathcal{P}}(f) \le \overline{S}_{\mathcal{P}}(g)$ and
$$
        \overline{\int_{a}^{b}}fdx \le \overline{S}_{\mathcal{P}}(f) \le \overline{S}_{\mathcal{P}}(g).
$$
Therefore,
$$
              \overline{\int_{a}^{b}}fdx \le \overline{S}_{\mathcal{P}}(g),
$$
and taking the infimum over all $\mathcal{P}$ gives
$$
             \overline{\int_{a}^{b}}fdx \le \overline{\int_{a}^{b}}gdx.
$$
A: From $f<g$ follows $0<g-f$ Hence,
Now integrate with same limits of integration:
$$\int 0 dx=0 < \int (g-f) dx = \int g dx - \int f dx$$
Or
$$\int f dx < \int g dx$$
