Additivity of definite integrals I'm trying to follow my textbook's prove for the following:
Let $f,g: [a,b] \rightarrow \mathbb{R} $ be integrable. Then $f+g$ is integrable and 
$ \int^{b}_{a}(f+g) = \int^{b}_{a}f + \int^{b}_{a}g $.  
The book basically shows that for any partition $P$, the lower sum $L(f+g,[a,b],P) = L(f,[a,b],P) + L(g,[a,b],P) $. This I understand.  
But from this it concludes, 
$ \underline\int^{b}_{a}(f+g) \leq \underline\int^{b}_{a}f + \underline\int^{b}_{a}g$ without any explanation.
Why is this the case? It seems to have skipped some steps.
 A: We have $\inf_{[a,b]} (f+g)\geq \inf_{[a,b]} f+\inf_{[a,b]} g$.
Now, let $\mathcal P$ be the set of partitions of $[a,b]$. Then, we have, for $P\in \mathcal P$, 
$L(f+g,P) \geq L(f,P) + L(g,P)$, 
There are partitions $Q$ and $R$ and their common refinement $P$ s.t.
$\underline{\int}_a^bf-\frac{\epsilon }{2} <L(f,Q)\leq L(f,P)$ and $\underline{\int}_a^bf-\frac{\epsilon }{2} < L(f,R)\leq L(f,P)$.
Then, $\underline{\int}_a^b(f + g)\geq L(f+g,P)\geq L(f,P) + L(g,P)\geq \underline{\int}_a^bf-\frac{\epsilon }{2}+\underline{\int}_a^bg-\frac{\epsilon }{2}=\underline{\int}_a^bf+\underline{\int}_a^bg -\epsilon$
from which it follows that $\underline{\int}_a^b(f + g)\geq \underline{\int}_a^bf+\underline{\int}_a^bg $, which is the desired inequality.
A: It's going to be impossible to prove that $f + g$ is integrable this way.
You ultimately want to show that
$$\underline{\int}_a^bf + \underline{\int}_a^bg \leqslant \underline{\int}_a^b(f + g) \leqslant  \overline{\int}_a^b(f + g) \leqslant \overline{\int}_a^bf + \overline{\int}_a^bg ,$$
in order to show that integrability of $f + g$ follows from that of $f$ and $g$.
Note that on any subinterval $I$ of a partition 
$$\inf_{x \in I}f(x) + \inf_{x \in I}g(x) \leqslant \inf_{x \in I}[f(x) + g(x)],$$
since for all $x \in I$,
$$f(x) + g(x) \geqslant \inf_{x \in I}f(x) + \inf_{x \in I}g(x).$$
Hence, forming lower sums we get,
$$L(f,P) + L(g,P) \leqslant L(f+g,P).$$
This is the correct starting point.
Assume that
$$\underline{\int}_a^b [f(x)+g(x)] \, dx < \underline{\int}_a^b f(x) \, dx + \underline{\int}_a^b g(x) \,dx ,$$
and show that this leads to a contradiction as follows. 
Assuming this were true we have
$$\underline{\int}_a^b [f(x)+g(x)] \, dx  -  \underline{\int}_a^b g(x) \,dx  <  \underline{\int}_a^b f(x) \, dx,$$
and there exists a partition $P$ such that
$$\underline{\int}_a^b [f(x)+g(x)] \, dx  -   \underline{\int}_a^b g(x) \,dx  <  L(P,f)  \leqslant \underline{\int}_a^b f(x) \, dx.$$
Hence,
$$\underline{\int}_a^b [f(x)+g(x)] \, dx  - L(P,f)  <   \underline{\int}_a^b g(x) \, dx,$$
and there exists a partition $P’$ such that
$$\underline{\int}_a^b [f(x)+g(x)] \, dx  - L(P,f)  <  L(P’,g) \leqslant \underline{\int}_a^b g(x) \, dx,$$
and
$$\underline{\int}_a^b [f(x)+g(x)] \, dx   <  L(P,f) + L(P’,g) .$$
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 
$$L(Q,f+g) \leqslant \underline{\int}_a^b [f(x)+g(x)] \, dx   <  L(P,f) + L(P’,g)  \leqslant L(Q,f) + L(Q,g).$$
This contradicts the inequality for lower sums, and, therefore
$$\underline{\int}_a^b f(x) \, dx + \underline{\int}_a^b g(x) \, dx \leqslant \underline{\int}_a^b [f(x) + g(x)] \, dx. $$
