The notion of improper Riemann integral is introduced precisely because the Riemann integral is not defined for unbounded functions and/or unbounded intervals. How are the upper and lower sums, $M(f,P)$ and $m(f,P)$, defined when the partition supposedly covers the unbounded interval $[d,\infty)$? There are special cases where infinite partitions and corresponding sums can be manipulated in a similar fashion to what is done for the true Riemann integral, but in general the machinery breaks down.
Focusing just on the integral over the semi-infinite interval $[a,\infty)$, if $f$ is Riemann integrable over $[a,c]$ for all $c > a$ then the improper integral is defined (pending existence) as
$$\int_a^\infty f(x) \, dx = \lim_{c \to \infty} \int_a^c f(x) \, dx.$$
Given that $\int_a^\infty |f(x)| \, dx$ exists, we have to prove that $\int_a^\infty f(x) \, dx$ exists. It is tempting to try immediately to show that the tail $\int_d^\infty f(x) \, dx$ can be made arbitrarily small by choosing sufficiently large d, but this is circular in that we have not yet established that an improper integral of $f$ over $[d,\infty)$ exists.
As with sequences and series, proof of existence without a prior candidate for the limit can be facilitated by a Cauchy criterion:
The improper integral exists if and only if for any $\epsilon > 0$
there exists $K > 0$ such that for all $c_2 > c_1 \geqslant K$ we have
$$\left|\int_{c_1}^{c_2}f(x) \, dx \right| < \epsilon$$
Armed with this theorem, the proof in question is straightforward.
Since the improper integral of $|f|$ exists, given $\epsilon > 0$ there exists $K$ such that with $c_2 > c_1 > K$ we have
$$ \tag{1} \left| \int_{c_1}^{c_2} f(x) \, dx\right| \leqslant \int_{c_1}^{c_2} |f(x)| \, dx < \epsilon.$$
Therefore, the improper integral of $f$ exists. Note that the first inequality in (1) is a well-known result for Riemann integrals.
Proof of Cauchy criterion:
Proof of the forward implication is straightforward. We are using the reverse implication here and the proof is as follows.
Define the sequence $I_n = \int_a^{a +n} f(x) \, dx $. From the hypotheses, given $\epsilon > 0$ there exists $K$ such that if the positive integers $m$ and $n$ satisfy $m > n \geqslant K- a$, then $a+m > a+n > K$ and
$$\tag{2}|I_m - I_n| = \left|\int_{a+n}^{a+m} f(x) \, dx \right| < \epsilon/2.$$
Hence, $(I_n)$ is a Cauchy sequence of real numbers and therefore converges to some real number $I$:
$$\tag{3}\lim_{n \to \infty} \int_a^{a+n} f(x) \, dx = I$$.
We can write
$$\tag{4}\left| \int_a^c f(x) \, dx - I \right| = \left| \int_{a+n}^c f(x) \, dx + \int_{a}^{a+n} f(x) \, dx - I \right| \\ \leqslant \left| \int_{a+n}^c f(x) \, dx \right| + \left| \int_{a}^{a+n} f(x) \, dx - I \right| .$$
If $c$ is sufficiently large, we can find $n$ sufficiently large with $c > a+n$, such that, using (2) and (3), the terms on the RHS of (4) are each less than $\epsilon/2$ . Therefore, the improper integral exists and takes the value $I$.
Your conjecture is true. Assume
$$\lim_{\substack{x \to -\infty \\ y \to \infty}}F(x,y)=I\in\Bbb R$$
and let $\epsilon>0.$ For some $N\in\Bbb N$ we have
$$\forall a,b\ge N\quad|F(-a,b)-I|<\epsilon$$
hence
$$\forall b,c\ge N\quad |F(0,b)-F(0,c)|=|F(-N,b)-F(-N,c)|<2\epsilon.$$
Therefore, by Cauchy's criterion,
$$ \lim_{y\to+\infty}F(0,y)$$
exists (in $\Bbb R$). Similarly (or by difference), so does
$$ \lim_{x\to-\infty}F(x,0).$$
Note that the existence of $ \lim_{y\to+\infty}F(z,y)$ and $\lim_{x\to-\infty}F(x,z)$ for some $z\in\Bbb R$ implies their existence for all $z\in\Bbb R.$
Best Answer
1.: Yes. The Improper Riemann integral is a limit of Riemann integrals, so $f$ must be Riemann integrable on $[a,b]$ for all $b>a$.
2.: Yes, you can suppose that $K_\varepsilon > a$ because you don't need to pick the smallest $K_\varepsilon$. And if $K_\varepsilon$ is good, then so is $42a+42K_\varepsilon$.
3.: The idea is good, but you can improve it. I would not integrate it from $a$ to $+\infty$, just from $a$ to some $b$, and examine the behaviour as $b \to +\infty$. Technically, you did the same, but I think it's more formal.
Also, the following line is not correct: $$\int_a^\infty f(x) \text{d}x =\int_a^{K_\varepsilon} f(x) \text{d}x+\int_{K_\varepsilon}^\infty f(x) \text{d}x > \int_a^\infty \frac{l}{2} \text{d}x =\infty$$ I think you wanted to write $$\int_a^\infty f(x) \text{d}x =\int_a^{K_\varepsilon} f(x) \text{d}x+\int_{K_\varepsilon}^\infty f(x) \text{d}x > \int_{\color{red}{K_\varepsilon}}^\infty \frac{l}{2} \text{d}x =\infty$$ But it is still not correct, because $$\int_a^{K_\varepsilon} f(x) \text{d}x$$ can be negative.