Sequence of measurable functions converging a.e. to a measurable function? I understand if $(X, \Sigma, \mu)$ is a measure space, and we have a sequence of measurable functions $f_{n}$ such that $\lim \limits_{n \to \infty} f_{n}$ exists almost everywhere d$\mu$ (a.e. d$\mu$), then it's equal to a measurable function almost everywhere.  The way this is constructed is to first call the set where the limit doesn't exist $N$ (and this clearly has measure $0$).  Then, we:
Define a new sequence 
$$\tilde{f_{n}} = \begin{cases} f_{n}(x) & x \not \in N \\
0 & x \in N \end{cases}$$
and we can think of $\tilde{f_{n}}$ as $f_{n} - f_{n}\chi_{N}$, where $\chi_{N}$ is the characteristic function of the set $N$.  Clearly, $\tilde{f_{n}}$ is measurable since it is the sum of measurable functions.  And so the limit of the sequence $\{ \tilde{f_{n}} \}$ is defined everywhere and measurable.  Specifically,
$$\lim \limits_{n \to \infty} \tilde{f_{n}} = \begin{cases} \lim \limits_{n \to \infty} f_{n} & x \not \in N \\ 0 & x \in N. \end{cases} $$
So we have a measurable function which is equal to $\lim \limits_{n \to \infty} f_{n}$ except on $N$.  But my question is:
$\lim \limits_{n \to \infty} f_{n}$ is only defined on $X \setminus N$, which means its domain is $X \setminus N$.  Why do we say this is equal to $\lim \limits_{n \to \infty} \tilde{f_{n}}$ almost everywhere if they have different domains?  Does it make sense to talk about them not being equal on $N$ if one of them isn't even defined on $N$?
I guess my question is:  If $X \subseteq X'$, and $f : X \to Y$ and $g : X' \to Y$, suppose $f = g$ on $X$ (with $X' \setminus X$ having measure $0$).  Does the statement $f = g$ a.e. even make sense?  We can't compare then on $X' \setminus X$ to say they aren't equal on it because one of the functions isn't even defined on that set.
Another question that has been spawned from this question is whether it makes sense to integrate a function over a set that is not in its domain.
 A: You can define two functions, one on the real numbers and one on the irrational numbers; you can say they are equal (or other properties) a.e with respect to the Lebesgue measure. 
Here is my understanding:
The main idea is that when you are doing integration, the Lebesgue integral will not see the difference of these two functions. The reason for this is because how Lebesgue integral is constructed, it looks at the pre-image $\{x\in \mathbb{R} : f(x) = a\}$. Note that this makes perfect sense even if $f$ is only defined on the irrational numbers.
Edit: to answer your question, I would suggest you to look at the construction of Lebesgue integral. In your example, the left hand side is perfectly defined. Here is an example, let $f:\mathbb{Q}^c \cap [0,1] \rightarrow \mathbb{R}$, $f\equiv 1$, and $g: [0,1] \rightarrow \mathbb{R}$, $g\equiv 1$; 
$$\int_0^1 f d\mu = 1\mu(\{x\in [0,1] : f(x) = 1\}) = 1*\mu(\mathbb{Q}^c \cap [0,1]) = 1 = \int_0^1 g d\mu.$$
The first equality above is from the definition of Lebesgue integral for characteristic function $\chi_A$.
Edit 2: Define $f:\mathbb{Q} \rightarrow \mathbb{R}$ with $f\equiv 1$. Now let us integrate over a more absurd space, $\mathbb{Q}^c$
$$\int_{\mathbb{Q}^c} f d\mu = 1\mu(\{x\in \mathbb{Q}^c : f(x) = 1\}) = 1\mu(\emptyset) = 0.$$
