About a proof that functions differing from a measurable function on a null set are measurable Consider the following problem: 

Say $f: M \to \mathbb{R} $ is measurable function, $M$ a measurable set and $g : M \to \mathbb{R}$ such that $ Y = \{ x : f(x) \neq g(x) \}  $ is a null set. We want to show $g$ is measurable.

So the solution would be to consider the difference $d = g - f$. This function $d$ is zero except on a null set.consider the following set 
$$
 \{x : d(x) > a \} =
 \begin{cases}
 N, & \text{if }a \geq 0 \\
 N^c, & \text{if } a < 0
 \end{cases}
 $$
Where $N$ is null set. Then it follows that $d$ is measurable and therefore $g$ is measurable. However, I still feel dirty on the set my teacher considered. Why $\{x : d(x) > a \}$ is equal to complement of null set $N$ when $a < 0 $ ?
 A: The expression that troubles you:
$$\{x: d(x) > a\} = \begin{cases}N&: a \ge 0\\N^c&: a < 0 \end{cases}$$
is really an abuse of notation. What is meant is something like:
$$\{x: d(x) > a\} = \begin{cases}N_a&: a \ge 0\\N_a^c&: a < 0 \end{cases}$$
because the null set $N_a$ varies with $a$, in the following way:
$$N_a = \begin{cases}\{x: d(x) > a\}&: a \ge 0\\ \{x: d(x) \le a\} &: a < 0\end{cases}$$
(This definition of $N_a$ is a consequence of the asserted equality above.)
The idea is that for each choice of $a$, we have that:
$$N_a \subseteq \{x: d(x) \ne 0\} = \{x: f(x) \ne g(x)\}$$
(check it!) and we know by assumption that the latter is a null set. So each $N_a$ is a subset of a null set. 
So, provided $N_a$ is actually measurable, it must have $\mu(N_a) = 0$ by monotonicity of the measure $\mu$. 
The way the solution is set up indicates that indeed each $N_a$ is measurable. Measure spaces having this property, where each subset of a null set is also a measurable set (and hence a null set), are called complete measure spaces.
Every measure space can be modified in a nonessential way to become a complete measure space; however the proof of this completion theorem for measure spaces is a bit technical so you may want to take it for granted as a justification for the assumption in your question.

Coming back to the original question, if we look at the case where $a <0$, we have that $\{x: d(x) > a\}$ is the complement of the null set $N_a = \{x :d(x) \le a\}$. It thus does not bear any relation to the null set $N_{a'} = \{x:d(x) >a'\}$ for any $a' > 0$.
You are therefore right to conclude that the solution as written is "dirty", because it obscures the fact that the null set $N$ depends on $a$ in a nontrivial way.
I hope this clarifies the matter for you.
