Form $w$ is integrable $\iff$ for all exhaustion $M$ there exists $\lim_{i\to \infty}\int_{K_i}w$ 
Let $M$ be a oriented non-compact surface of dimension $m$ and $w$ be
  a non-negative $m$-form continuous on $M$. An exhaustion of $M$ is a
  sequence of compact $J$-measurable $K_i$ such that $M=\cup K_i$ and
  $K_i\subset int K_{i+1}$ for $i=1,\cdots$. Prove that $w$ is
  integrable $\iff$ for all exhaustion $M$ there exists $\lim_{i\to
 \infty}\int_{K_i}w$

According to my book, $\int_M w$ is defined when $w$ is an $m$-form of compact support in $M$, following these two conditions:
1) The set of points of $M$ in which $w$ is discontinuous has null measure
2) $w$ is bounded, that is, there is $c>0$ such that 
$$|w(x)\cdot (w_1,\cdots, w_m)|\le c|w_1|\cdots|w_m|$$
for any $x\in M$ and $w_1,\cdots,w_m\in T_xM$
If $f$ is integrable, then it's bounded and its set of discontinuity has measure $0$. The last condition may say something about the integral, but I'd have to use the definition of integrability again to say that $\lim_{i\to\infty}\int_{K_i}w$ exists. Why does $\int_{K_i}w$ exists to begin with? Because $K_i$ is compact? 
Now, if $\lim_{i\to\infty}\int_{K_i}w$ exists, then what does it mean exactly? $|\int_{K_i}w-L|<\epsilon$ for any $\epsilon$ when $i> N$, but... how does it help?
UPDATE:
I think made some progress on $\leftarrow$:
If for all exhaustion $M$ there exists $\lim_{i\to
 \infty}\int_{K_i}w$, it means that the sequence $\int_{K_i}w$ is bounded, which means that there exists $M$ such that $\int_{K_i}w< M$ for all $i$, right? Shouldn't I be able to prove that $w$ is bounded? Don't I need the integral to exist for at least one $i$? What about an argument on the set of discontinuity?
 A: First, note that $w$ is assumed to be continuous. This means that condition 1 is automatically satisfied (the empty set has null measure), and condition 2 holds on compact sets (because continuous functions are bounded on compact sets). In particular, this means that the integral $\int_K w$ is well-defined and finite for any $J$-measurable compact set $K \subset M$. Moreover, if $w$ has compact support, then $\int_M w$ is also well-defined and finite, i.e. $w$ is integrable.
A second useful observation is that the sequence $\{ \int_{K_i} w \}_{i \ge 1}$ is increasing (here I am using the fact that $w$ is non-negative). This implies that it converges if and only if it is bounded. Hence, it suffices to prove the following.

Proposition: $w$ is integrable if and only if the exists a constant $C$ such that $\int_K w \le C$ for all $J$-measurable compact sets $K \subset M$.

If $w$ has compact support, then as noted above it is automatically integrable, and we can take $C = \int_M w$. Thus the proposition is trivially true in this case.
Things get more interesting if $w$ is not compactly supported. In general, the integral of a non-compactly supported form is not well-defined. However, since the form $w$ we are considering is non-negative, we can define its integral to be
$$ \int_M w = \sum_i \int_M \rho_i w $$
where $\{ \rho_i \}$ is a partition of unity on $M$. For each $i$, the form $\rho_i w$ is continuous and has compact support, so the above sum makes sense. Moreover, all the terms are non-negative, so there are only two possibilities: either the series converges absolutely to a real number, or it is infinite. If the former holds, we say that $w$ is integrable. One can use the properties of absolutely convergent series to show that the value of the integral $\int_M w$ does not depend on the partition of unity chosen.
We are now ready to prove the proposition.
$(\Rightarrow)$ Since $w$ is non-negative and integrable, we have $\int_K w \le \int_M w < \infty$ for all $J$-measurable compact $K \subset M$, so we can take $C = \int_M w$.
$(\Leftarrow)$ Let $\{\rho_i\}_{i \in I}$ be a partition of unity on $M$. For any finite subset $J \subset I$, we have
$$ \sum_{j \in J} \int_M \rho_j w = \int_M \left(\sum_{j \in J} \rho_j \right) w \le C $$ since the support of the function $\sum_{j \in J} \rho_j$ is compact. This shows that the partial sums of the series $\sum_i \int_M \rho_i w$ are bounded and hence that the series converges, so that $w$ is integrable.
