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.
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.