Is there a difference between those two notions of "almost everywhere"? Assume we have two functions $f$ and $g$ on, say, the interval $[0,1]$. Let's say that $g$ is continuous, i.e. the pointwise evaluation $g(x)$ makes sense. $f$, on the other hand, is only $L^1([0,1])$ and thus not defined pointwise.
Is there a difference between saying 
a) $f = g$ a.e. on [0,1]$ and
b) There is a fixed negligible set $N$ such that $f(x) = g(x)$ for all $x\not \in N$?
On first glance it seems that b) is only the expanded definition of a) (as defined, e.g., on Wikipedia), but isn't b) stronger than a)? 
I would reason that in the setting of b), for any point outside $N$ we can now give $f(x)$ proper meaning (as being defined by $g(x)$), while in the setting a) we can never tack down any pointwise evaluation because we can always change any arbitrary point's image $f(x)$ to any value with a) still being valid.
If there is indeed a difference, is there a name for b) (or a shorter way of writing it)? 
 A: I think I may understand the confusion.  There are really two meanings to $f = g$ a.e., by abuse of notation.  The first is your (b): there is some set $N$ of measure $0$ outside of which $f=g$.  Here $f(x)$ and $g(x)$ are still defined at any point $x$.
The second meaning of $f=g$ a.e. is where $f$ and $g$ are taken to be equivalence classes.  The elements of $L^p$ are often taken to be equivalence classes of functions, where $f \sim g$ if $f = g$ a.e.  This is convenient because, e.g., we can define $||f||_p = (\int |f|^p)^{1/p}$ as a norm - part of the definition of a norm $||\cdot||$ is that $||f|| = 0$ implies $f = 0$.  So when we write $f=g$ a.e. in this context, it really just means that $f=g$ as equivalence classes.  But I think some authors will still write $f=g$ a.e. If we are considering $f$ and $g$ to be equivalence classes of functions rather than functions, it does not make sense to evaluate $f$ and $g$ at any point $x$ (unless it is an atom, with $\mu(\{x\}) > 0$.)  Nevertheless sometimes the notation is abused, with authors writing things like $f(x) = x^2$ has $f \in L^1[0,1]$.  Basically, the equivalence relation means everything in $L^p$ is just understood to only matter up to sets of measure zero.
