I'm trying to understand a proof about lower and upper Riemann integrals that has been covered during my lecture, but I'm stuck at the conclusion, since I'm apparently missing something about the use of $\epsilon$. It's probably something very stupid and obvious, but I still don't get it.
We want to prove that for all bounded functions $f, g : [a,b] \rightarrow \mathbb{R}$, we have: $$\int_a^{b*} (f+g)(x) dx \le \int_a^{b*}f(x)dx + \int_a^{b*}g(x)dx$$
where $$\int_a^{b*}f(x)dx := \inf\left\{ \int_a^b \psi(x) dx : \psi \in T([a,b]), \psi \ge f \right\} $$ and $T([a,b])$ is the set of all step functions $h:[a,b] \rightarrow \mathbb{R}$.
The proof is the following: (I'll write it exactly as it is written in my lecture)
Let $\epsilon > 0$. $\Rightarrow \exists$ $\psi_0$, $\psi_1 \in T([a,b])$ such that $\psi_0 \ge f$, $\psi_1 \ge g$ and
$$\int_a^b \psi_0 (x) dx \le \int_a^{b*}f(x) dx + \frac{\epsilon}{2}$$ $$\int_a^b \psi_1 (x) dx \le \int_a^{b*}g(x) dx + \frac{\epsilon}{2}$$
$\Rightarrow \psi:= \psi_0 + \psi_1 \in T([a,b])$ and $\psi = \psi_0 + \psi_1 \ge f+g$ (because of a lemma we covered before) $$\Rightarrow \int_a^{b*} (f+g)(x)dy \le \int_a^b \psi(x) dx = \int_a^b \psi_0(x) dx + \int_a^b\psi_1(x)dx$$ $$\le \int_a^{b*}f(x)dx + \int_a^{b*}g(x)dx + \epsilon$$
So far I think I understood everything. But here is the last sentence of the proof, and that's what I don't understand:
arbitrary $\epsilon > 0$ $\Rightarrow$ $\int_a^{b*} (f+g)(x) dx \le \int_a^{b*}f(x)dx + \int_a^{b*}g(x)dx$.
...and I feel so sorry while writing this because I know how stupid it is to not understand such a thing, but here is what I don't get: we have:
$$\int_a^{b*}(f+g)(x)dx \le \int_a^{b*}f(x)dx + \int_a^{b*}g(x)dx + \epsilon$$
And let's say that the left part of that thing is $A$, and the right part is $B+\epsilon$. So we have $A \le B+\epsilon$.
We want to show that $A \le B$.
But all we know is that: $$A-\epsilon \le A \le A+\epsilon$$ $$B-\epsilon \le B \le B+\epsilon$$ $$A \le B+\epsilon \text{ (so } A-\epsilon \le B)$$
So why should we have automatically $A\le B$, even for a very small $\epsilon$? Why couldn't we have $B \le A$? What does "arbitrary $\epsilon > 0$" mean? Shouldn't we have the result for any $\epsilon > 0$, not only small ones?
Again, I'm really sorry if my question sounds stupid. I already looked for other posts on stackexchange about that subject but nothing really convinced me. I found this post: The method of proving the equality of integrals by showing they agree within $\epsilon$, for an arbitrary $\epsilon>0$ and I tried to understand the theorem that Sudarsan mentioned but I felt like the theorem was wrong (and I've not been able to prove it... I'm really going to lose my mind on that thing.)
So... any help would be greatly appreciated! Thank you very much in advance. :)
EDIT: After some more hours of research, I managed to find a "proof" (which, I hope, is correct) for the following statement: Let $a, b, c \in \mathbb{R}$. If $c>a$ for all $c>b$, then $a \le b$. But I'm still upset about the $\le$ that appears in the precedent proof, and I'm still confused about all this. I'll try to give more details tomorrow, especially if there's still no answer or comment until then.
Since $A \leq B + \epsilon$ for any $\epsilon>0$, it must be true that $A \leq B$. Otherwise, just take any $\epsilon >0$ such that $\epsilon < (A-B) $. This particular $\epsilon$ satisfies $B + \epsilon < A$, which is a contradiction.