Difficulty with Proof of Integrability I'm working on a proof on integrability, and cannot seem to grasp the final step.
The theorem is that: if we have two integrable functions, $f$ and $g$, over the interval $[a,b]$, then we can say their sum, $f + g$, is also integrable and equal to the sum of their integrals. I can prove everything but the final step: that $\int_a^b (f + g) = \int_a^b f + \int_a^b g$. 
I understand, I believe, everything up until the point that we conclude $U(f+g, P) - L(f+g, P) < \epsilon$, and thus $f+g$ is integrable. From here, there seems to be a jump into concluding that, thus the integral of $f + g$ is equal to the sum of the integral of $f$ and $g$, and seems to have something to do with the definition of integrability, i.e., since it is integrable, we can conclude that $\int (f+g) = U(f+g) = L(f+g)$, and we had already that $\int f = L(f) = U(f)$ and $\int g = L(g) = L(f)$, but I can't figure out how to string these facts together.
Any help would be greatly appreciated. 
 A: To avoid confusion, let's use the standard notation and definitions for lower and upper sums and integrals:
$$\overline{\int}_a^b f = \inf_P U(P,f), \quad\underline{\int}_a^b f = \sup_P L(P,f). $$
If we can show 
$$\tag{1}\underline{\int}_a^b f + \underline{\int}_a^b g \leqslant \underline{\int}_a^b (f+g) \leqslant \overline{\int}_a^b (f+g) \leqslant \overline{\int}_a^b f + \overline{\int}_a^b g, $$
we are finished, since if $f$ and $g$ are integrable both the LHS and RHS equal $\int_a^b f + \int_a^bg$.  This forces equality between the lower and upper integrals of $f+g$, proving both integrability of $f+g$ and $\int_a^b(f+g) = \int_a^b f + \int_b g.$
I will show how to prove the inequality on the LHS of (1) by contradiction. The proof of the RHS inequality is similar.
If we assume 
$$\underline{\int}_a^b (f+g)  <  \underline{\int}_a^b f  +  \underline{\int}_a^b g ,$$
then it follows that 
$$\underline{\int}_a^b (f+g)  -  \underline{\int}_a^b g  <  \underline{\int}_a^b f,$$
and there exists a partition $P$ such that
$$\underline{\int}_a^b (f+g)  -   \underline{\int}_a^b g  <  L(P,f)  \leqslant \underline{\int}_a^b f.$$
Rearranging we get
$$\underline{\int}_a^b (f+g)  - L(P,f)  <   \underline{\int}_a^b g,$$
and there exists a partition $P’$ such that
$$\underline{\int}_a^b (f+g)  - L(P,f)  <  L(P’,g) \leqslant \underline{\int}_a^b g.$$
Hence,
$$\underline{\int}_a^b (f+g)   <  L(P,f) + L(P’,g) .$$
Now take a common refinement of the partitions $Q = P \cup P'$.  Lower sums increase as partitions are refined and we have $L(Q,f) \geqslant L(P,f)$ and $L(Q,g) \geqslant L(P’,g).$
It follows that 
$$\tag{2}L(Q,f+g) \leqslant \underline{\int}_a^b (f+g)   <  L(P,f) + L(P’,g)  \leqslant L(Q,f) + L(Q,g).$$
However, over any partition subinterval $I$ we have $\inf_{x \in I}f(x) + \inf_{x \in I}g(x) \leqslant \inf_{x \in I} (f+g)$ and it follows that 
$$\tag{3}L(Q,f) + L(Q,g) \leqslant L(Q,f+g),$$
which contradicts (2).
Therefore,
$$\tag{4}\underline{\int}_a^b f + \underline{\int}_a^b g \leqslant \underline{\int}_a^b (f+g). $$
Note
A common mistake is to start with (3) and assume it implies (4) directly since
$$\sup_Q [L(Q,f) + L(Q,g)] \leqslant \sup_Q L(Q,f+g). $$
The problem is the only additional information that can be inferred directly from the definition of the supremum is
$$\sup_Q [L(Q,f) + L(Q,g)] \leqslant \sup_Q L(Q,f) + \sup_Q L(Q,g), $$
and we cannot determine the order relationship between the right-hand sides of the two inequalities. 
