Extended integral over parabola open set 
Let $f(x,y)=1/(y+1)^2$ for all $x,y\in\mathbb{R}$. Let $A$ be the open set $$A=\{(x,y)\mid x>0, x^2<y<2x^2\}.$$ Show that $\int_A f$ exists, and calculate it.

Since $A$ is an unbounded open set, we have to use the "extended integral", which has two equivalent definitions.
1) $\int_A f$ is the supremum of the numbers $\int_D f$, as $D$ ranges over all compact Jordan-measurable subsets of $A$.
2) If $C_1,C_2,\ldots$ is a sequence of compact Jordan-measurable subsets of $A$ such that $\cup_{i=1}^{\infty}C_i=A$ and $C_i$ belongs to the interior of $C_{i+1}$ for all $i$. Then $\int_A f=\lim_{i\rightarrow\infty}\int_{C_i}f$.
Definition 2) is easier to deal with. Nevertheless, for a set like the set $A$, with a really strange shape, it is really hard to imagine what compact, rectifiable subsets we should take to satisfy the conditions in 2). Also, it must be easy to calculate the integrals of $f$ over these subsets, so that we can compute the limit to be the integral of $f$ over $A$.
Which subsets $C_1,C_2,\ldots$ should be chosen to satisfy the conditions in  2), so that we can calculate the integrals over them?
 A: We want a normal exhaustion of $A$ with Jordan measurable compacta $C_k$. To keep them compact, we must bound the $x$-coordinate away from $0$, and of course from infinity. To have $C_k$ contained in the interior of $C_{k+1}$, the bounds must be strictly monotonic. $\frac1k \leqslant x \leqslant 2k$ achieves that. Also, we must keep the $y$-coordinate away from the graphs of $y = x^2$ resp. $y = 2x^2$, that is achieved by adding resp. subtracting a suitable small constant $\varepsilon_k > 0$ (which we determine later). So let
$$C_k = \left\lbrace (x,y) \in \mathbb{R}^2 : \frac1k \leqslant x \leqslant 2k, x^2 + \varepsilon_k \leqslant y \leqslant 2x^2 - \varepsilon_k \right\rbrace.$$
To have $C_k \subset \operatorname{int} C_{k+1}$, we need $\varepsilon_{k+1} < \varepsilon_k$, and to have $A = \bigcup\limits_{k=1}^\infty C_k$, we need $\varepsilon_k \to 0$. To have $C_k$ behave nicely on the left boundary, we want
$$\frac{1}{k^2} + \varepsilon_k \leqslant \frac{2}{k^2} - \varepsilon_k \iff \varepsilon_k \leqslant \frac{1}{2k^2}.$$
Let's choose $\varepsilon_k = \frac{1}{4k^2}$. Then all conditions are met, $C_k$ is compact, $C_k \subset \operatorname{int} C_{k+1}$, $A = \bigcup\limits_{k=1}^\infty C_k$, and all $C_k$ are Jordan measurable since $C_k$ is the area between the graphs of two continuous functions defined on a closed interval (and closed intervals are Jordan measurable).
It remains to compute $\int_{C_k} \frac{1}{(y+1)^2}$.
$$\begin{align}
\int_{\frac1k}^{2k}& \left(\int_{x^2 + \varepsilon_k}^{2x^2-\varepsilon_k}\frac{1}{(y+1)^2}\,dy \right)\,dx\\
&= \int_{\frac1k}^{2k} \left( \frac{1}{x^2+\varepsilon_k+1} - \frac{1}{2x^2-\varepsilon_k+1}\right)\,dx\\
&= \frac{1}{\sqrt{1+\varepsilon_k}}\left(\arctan \left(\frac{2k}{\sqrt{1+\varepsilon_k}}\right) - \arctan\left(\frac{1}{k\sqrt{1+\varepsilon_k}}\right)\right)\\
&\quad -\frac{1}{\sqrt{2(1-\varepsilon_k)}}\left(\arctan\left(\frac{2\sqrt{2}k}{\sqrt{1-\varepsilon_k}}\right) -\arctan\left(\frac{\sqrt{2}}{k\sqrt{1-\varepsilon_k}}\right)\right).
\end{align}$$
The limit for $k\to \infty$ is
$$\frac{\pi(\sqrt{2}-1)}{2\sqrt{2}}.$$
A: O.k, I'm not completely sure this is correct, but here goes. 
If we take the Jordan measurable sets as slices of $\bf A$ going between $x=0$ and $x=n$, so that that region is your $C_n$ (this is the bit I'm not sure of - think these are measurable but don't know how to prove it? Never was much good at measure theory), then we can write $\int_{\bf A} f$ as
\begin{equation}
\lim_{n \to\infty} \int_0^n \int_{y=x^2}^{y=2x^2} \frac{1}{(y+1)^2} \mathrm{d}y\mathrm{d}x
\end{equation}
Which we can evaluate exactly - 
\begin{align}
=\int_0^n (\frac{-1}{2x^2+1} + \frac{1}{x^2 +1}) \mathrm{d}x\\
=[\mbox{arctan}(x) - \frac{1}{\sqrt 2} \mbox{arctan
} (\sqrt{2} x) ]_0^n
\end{align}
Evaluating this at $0$ gives no contribution and so we're left with the whole integral being equal to
\begin{equation}
\lim_{n \to \infty}\,\, (\mbox{arctan} (n)  - \frac{1}{\sqrt 2} \mbox{arctan} (\sqrt{2} n)) = \frac{\pi (\sqrt{2}-1)}{2\sqrt{2}}
\end{equation} 
Definitely playing fast and loose here so if someone wants to correct me or make this more rigorous, be my guest. Think it makes sense though. 
