A (partial) argument converting sums in $\ell^1$ into Lebesgue-integrable functions. First, I want to mention that this problem is off of a take home final I have. I was given permission to research/ask about this specific line of reasoning, in large because I think the professor thinks it might work and this is not a ``standard'' solution to this problem.  
The question on the exam is:

Given  a measurable space $(X,B)$, and $\mu_n$ a sequence of measures on $(X,B)$ with $\mu_n(A) \rightarrow \mu(A)$ for each $A \in B$. If $\mu(X) < \infty$, then $\mu$ is a measure. 

A little notation: Let $\{a^{(n)}\}$ be a collection of sequences, where the $i$th term in the $j$th sequence is given by $a^{(j)}_i$. The above theorem can be deduced readily from the following lemma:

Let $\{a^{(n)}\}$ be a collection of sequences of non-negative real numbers. Suppose $\exists M >0$ such that $\sum_{i=1}^\infty a^{(n)}_i < M$ for each $n$. Suppose further that $a^{(n)}_i \rightarrow a_i$, some real number as $n \rightarrow \infty$. Then $$\lim_{n \rightarrow \infty} \sum_{i=1}^\infty a^{(n)}_i =\sum_{i=1}^\infty a_i$$

Now, my attempt at proving the lemma was to turn this into a problem in measure theory. Given $\{a^{(n)}\}$ as in the lemma, define $$\phi_n(x) = \sum_{i=1}^\infty a^{(n)}_i \chi_{(i,i+1]}$$.
Where $\chi_A$ is the usuall characteristic function for a set $A$.  Intuitively, $\phi_n(x)$ looks like an infinite step function, where $\phi_n(x) = a^{(n)}_i$ if $x \in (i,i+1]$. Now, it is clear that $\int \phi_n(x) dx = \sum_{i=1}^\infty a^{(n)}_i$. Moreover, we have that $\phi_n(x) \rightarrow \phi(x) := \sum_{i=1}^\infty a_i \chi_{(i,i+1]}$. 
All that remains is to show:
$$\lim_{n\rightarrow \infty} \int \phi_n(x) = \int \phi(x)$$
This will require some kind of convergence theorem. There doesn't seem to be one readily available, but these functions have lots of special properties. Such a convergence theorem would start with ``Let $\{f_n\}$ be a sequence of non-negative Lebesgue integrable functions. '' Here are some hypotheses one could take:


*

*$f_n \rightarrow f$ pointwise.

*$\int f_n  < M$ for some universal bound $M$.

*As a sequence in $n$, $\int f_n$ converges and is finite.

*$\|f_n\|_\infty < M$ for some universal constant $M$.


Unfortunately, I don't believe these hypotheses quite do it. Say $\phi_n = \chi_{(n,n+1]}$. These guys converge pointwise to $0$ but in integral to $1$. My question is: Is there a set of additional hypotheses one can take to deduce a convergence theorem that resolves the lemma? 
EDIT: Okay, it has been pointed out that the lemma isn't true (with a clear counterexample). The question now becomes can one take the theorem, and build the correct lemma that allows you to employ this kind of technique? I've basically managed to do nothing but persuade myself that $\mu(X)< \infty$ must tell you more than  one can assume there is some $M$ such that for any countable collection of disjoint sets in $B$ $\sum \mu_n(A_i) < M$
 A: Here a few observations.
As I stated in the comments, your lemma basically seems to be you can change the order of summation and taking the limit. Since the summation is also a limit, that's the same as changing the order of two limits, which in general requires uniform convergence of the inner limit in respect to the parameter of the outer limit. The counter-example of David Mitra exploits exactly that - if $a^{(n)}$ is the $n$-th unit vector, the sum doesn't converge uniformly in respect to $n$, and thus the lemma fails.
In your attempt to prove the lemma, you end with trying to prove $$
 \lim_{n\to\infty} \int \phi_n = \int \phi \text{.}
$$
You get that from the dominated convergence theorem I if there's an integrable function $\Phi$ with $\phi_n(x) < \phi(x)$ for all $x$. You don't have that either, though, in the case of the unit vectors, since $\Phi(x)$ would have to be $\geq 1$ on all intervals $[i,i+1)$, which makes it non-integrable. In fact, if there was such a $\Phi$, then it'd give you a bound on $s_k = \sum_{i=k}^\infty a^{(n)}_i$ independent of $n$, which would also give you uniform convergence of $\sum_{i=0}^\infty a^{(n)}_i$.
Let's see what happens if we try to translate the unit-vector counter-example into measures and sets. Let $X = \mathbb{R}$ and let $\mu_n(A) = 1$ if $n\in A$, $0$ otherwise. Set $A_i = [i,i+1)$. Then $\mu_i(A_j) = \delta_{i,j}$ ($\delta_{i,j}$ is the Kronecker delta). $\mu(A_i) = 0$, but $\mu(\bigcup_i A_i)=1$. Wait...
So whats' going wrong here? Does this disprove the theorem? It doesn't, because it doesn't really fullfill $\mu_n(C) \to \mu(C)$ for every $C$. Let for example $C = \{1,3,5,\ldots\}$. Then $\lim_{n\to\infty}\mu_n(C)$ doesn't even exist (it oscilates between $0$ and $1$). By understanding why this happens, one might be able to come up with a way to fix your approach, but I'm not there yet..
