Show that $\overline{\int_a^b} f+g \le \overline{\int_a^b} f+\overline{\int_a^b}g$. Show that $\overline{\int_a^b} f+g \le \overline{\int_a^b} f+\overline{\int_a^b}g$. 
Attempt: Using the fact that $\sup(x+y)\le \sup(x)+\sup(y),$ 
$$\inf (U(P,f+g))\le \inf(U(P,f)+U(P,g))$$ for any partition $P$ on $[a,b]$.
I am a little bit confused here. Is $\inf(U(P,f)+U(P,g))=\inf(U(P,f))+\inf(U(P,g))?$
 A: To prove by contradiction, assume that
$$\overline{\int_a^b} (f + g)  >  \overline{\int_a^b} f  +  \overline{\int_a^b} g.$$
Then there exists a partition $P$ such that
$$\overline{\int_a^b} (f+g)  -   \overline{\int_a^b} g  >  U(P,f)  \geqslant \overline{\int_a^b} f.$$
Since,
$$\overline{\int_a^b} (f+g)  - U(P,f)  >   \overline{\int_a^b} g,$$
there exists a partition $P’$ such that
$$\overline{\int_a^b} (f+g)  - U(P,f)  >  U(P’,g) \geqslant \overline{\int_a^b} g,$$
and
$$\overline{\int_a^b}(f+g)   >  U(P,f) + U(P’,g) .$$
Take a common refinement of the partitions $Q = P \cup P'$. Since upper sums decrease as partitions are refined, we have,
$$\tag{*}U(Q,f+g) \geqslant \overline{\int_a^b} (f+g)   >  U(P,f) + U(P’,g)  \geqslant U(Q,f) + U(Q,g).$$
However, $\sup [f(x) + g(x)] \leqslant \sup f(x) + \sup g(x)$ and it follows that  
$$U(Q,f+g) \leqslant U(Q,f) + U(Q,g),$$ 
which contradicts (*).
A: No, but you can prove the result using common refinement. Let $\epsilon > 0$ and choose partitions $P$ and $Q$ such that $$U(P,f) < \overline{\int_a^b} f + \frac{\epsilon}{2}\quad \text{and}\quad U(Q,g) < \overline{\int_a^b} g + \frac{\epsilon}{2}$$
If $T$ be the common refinement of $P$ and $Q$, show that 
$$U(T,f+g) \le \overline{\int_a^b} f + \overline{\int_a^b} g + \epsilon$$
using $U(T,f) \le U(P,f)$ and $U(T,g)\le U(Q,g)$. Then 
$$\overline{\int_a^b} (f + g) \le \overline{\int_a^b} f + \overline{\int_a^b} g + \epsilon$$
Since $\epsilon$ was arbitrary, the result follows.
