Proving sum of integrals is integral of sums. This comes from Spivak "Calculus on Manifolds" problem 3-3. I think I have proof but it seems too simple to be true so I would like others to look through it and maybe point me to some subtle issues. 
I want to prove that if $f$ and $g$ are integrable functions $A \to \mathbb{R}$ where $A$ is a closed rectangle in $\mathbb{R}^n$, then $f+g$ is also integrable. Also, I would like to prove that the values are related in the following way.
$$ \int_A f+g = \int_A f + \int_A g$$
I assume that the following has already been proven. Here, $L(f,P)$ is lower sum of $f$ for partition $P$, $U(f,P)$ is upper sum of $f$ for partition $P$.
$$ L(f,P) + L(g,P) \leq L(f+g,P) $$
$$ U(f,P) + U(g,P) \geq U(f+g,P) $$
Then, because $f$ and $g$ are integrable, then for every $\varepsilon/2 > 0$ we can find partitions $P$ and $P'$ such that the following is true.
$$ U(f,P) - \frac{\varepsilon}{2} \leq \int_A f \leq L(f,P) + \frac{\varepsilon}{2}$$
$$ U(g,P') - \frac{\varepsilon}{2} \leq \int_A g \leq L(g,P') + \frac{\varepsilon}{2} $$
Combining results, we get the following.
$$ U(f,P) + U(g,P') - \varepsilon \leq \int_A f+\int_A g \leq L(f,P)+L(g,P')  +\varepsilon $$
Now, take refinement of $P$,$P'$ and call it $P''$. Then we have the following results for $f$ and similarly for $g$.
$$ U(f,P'') \leq U(f,P) $$
$$ L(f,P'') \geq L(f,P) $$
So, combining multiple results, we have the following inequalities.
$$ U(f+g, P'') - \varepsilon \leq \int_A f+ \int_A g \leq L(f+g,P'') + \varepsilon$$
From here, first of all, we see that $f+g$ is integrable.
$$ U(f+g, P'') - L(f+g, P'') \leq 2 \varepsilon $$
Now, consider $\mathrm{sup}(L(f+g,P'')) = \alpha$. Assume that $\alpha < \int_A f+ \int_A g $. 
But then take $\varepsilon = (1/2)(\int_A f+ \int_A g - \alpha) > 0$. In that case, the following inequality shows that it contradicts that $\alpha$ is an upper bound.
$$ \alpha < \frac{\alpha + \int_A f + \int_A g}{2} = -\frac{\int_A f + \int_A g - \alpha}{2} + \int_A f + \int_A g \leq L(f+g,P'') $$
Assume that $\alpha > \int_A f + \int_A g$. But then take $\varepsilon = (1/2)(\alpha-\int_A f - \int_A g) > 0$. In that case, the following inequality shows that it contradicts that $\alpha$ is the least upper bound. This is due to each upper sum is an upper bound for the set of lower sums.
$$ \alpha > \frac{\alpha + \int_A f + \int_A g}{2} = \frac{\alpha - \int_A f - \int_A g}{2} +\int_A f + \int_A g \geq U(f+g,P'') $$
I would really appreciate your thoughts and comments.
 A: The proof looks fine. One thing to note is that a few pages later, Spivak proves the theorem that a function $f: A \to \Bbb{R}$ ($A \subset \Bbb{R}^n$ being a closed rectangle) is Riemann integrable if and only if the set of discontinuities of $f$ has (Lebesgue) measure zero in $\Bbb{R}^n$. Using this theorem, the proof that $f+g$ is integrable can be made considerably shorter.

Your proof for $\int_A(f+g) = \int_Af + \int_A g$ looks fine, but here's a rephrasing of it which I think is quicker (if only very slightly so).
Let $\epsilon > 0$ be arbitrary. You've shown that there exists a partition $P''$ such that
\begin{align}
U(f+g, P'') - \epsilon \leq \int_Af + \int_Ag \leq L(f+g,P'') + \epsilon.
\end{align}
Now, note that pretty much by definition, we also have:
\begin{align}
L(f+g,P'') \leq \int_A(f+g) \leq U(f+g, P'').
\end{align}
Subtracting the inequalities, we find that
\begin{align}
- \epsilon \leq \int_Af + \int_Ag - \int_A(f+g) \leq \epsilon.
\end{align}
Since $\epsilon > 0$ was arbitrary, it follows that the quantity in the middle has to be $0$; hence $\int_A f + \int_A g = \int_A(f+g)$.
Of course, here, I'm pretty much invoking the familiar fact that if $x$ is a number such that for all $\epsilon > 0$, $|x| \leq \epsilon$, then $x=0$ (for which the proof is of course pretty much what you said; if $x \neq 0$, take $\epsilon = \frac{|x|}{2}$ and obtain a contradiction). This simply "avoids" the double case work which you did.
