Limit of an integral over a sequence of compact sets I was reading an old calculus book today(Functions of Several Variables by Wendell Fleming) and I came across an exercise I am having a great deal of trouble completing:

let $f$ be continuous on an open set $D$. Assume that the integrals of $f^+$ and $f^-$ over $D$ both diverge to $+\infty$. Show that given any real number $l$ there is a sequence of compact sets $K_1 \subset K_2 \subset ...$ such that $D = K_1 \cup K_2 \cup...$ and $lim_{ i \to \infty} \int_{K_i} fdV = l$.

Any help would be appreciated, either in the form of a hint or an actual proof. It does not seem easy. Thank you.
For clarity $f^+(x) = \max \{f(x), 0\}$ and $f^-(x) = \max \{-f(x), 0\}$.
 A: We can reduce it to the following
Lemma: Let $f$ be continuous on an open set $D$ with integrals of $f^+$ and $f^-$ diverging. Then, for any $l\in\Bbb R$, there is a compact set $K$ with $\int_Kf=l$.
At least if $D\subseteq\Bbb R$, it follows from the continuity of the integral function. 
Similar arguments may work for $D\subseteq\Bbb R^n$.

Then, take an arbitrary sequence $a_1,a_2,\dots\to l$, and choose $K_1$ by the lemma for $l_1:=a_1$ and $D_1:=D$,
then take $D_{n+1}:=D_n\setminus K_n$ still open, and the integral of $f$ reduced only by the finite $l$, and $l_{n+1}:=a_{n+1}-a_n$.
A: First look at when $D$ is bounded, fix $N$ such that 
$$\int_{\{0\leq f\leq N\}} f >l$$ 
we see that for each
$n\geq N$, there exists $a_n$ strictly decreasing such that 
$$\int_{\{0\leq f\leq n\}} f + \int_{\{a_n \leq f\leq 0\}} f = l.\quad\quad\quad(\star)$$
This comes from the fact that $g(y) =\int_{\{y \leq f\leq 0\}} f$ is a continuous monotone decreasing function $g:[0,-\infty) \rightarrow [0,-\infty)$ and using intermediate value theorem for $$g(a_n) = l-\int_{\{0\leq f\leq n\}} f.$$  
Since we assumed $D$ is bounded, we see that $a_n$ can not be bounded below, if they are, this implies the second integral in $(\star)$ is bounded, but the first one is not. Thus we can define 
$$K_n =  \{0\leq f\leq n\} \cup \{a_n \leq f\leq 0\}.$$
We have $\int_{K_n} f = l$ for all $n\geq N$, $K_n$'s are compact and they are monotone increasing sets with their union being $D$.
This part is not correct:
Now for general $D$, rewrite it as union of bounded disjoint open sets $\{D_i\}$. From above result, you will get a sequence of $\{K_{i,n}\}$ for each $D_i$, and you can define 
$$K'_m = \cup_{i=1}^m \cup_{n=1}^m K_{i,n}.$$
