Question About Definition of Almost Everywhere I suppose I'm a bit confused about the definition in the following regard:
A property holds a.e. if it holds everywhere except for a set of measure $0$. Now, if the particular property is only defined for a set of measure $0$, is it a.e. by default?  
Say I have two 'continuous' (standard topology) sequences $f,g: \mathbb{N} \to \mathbb{R}.$ Are we then allowed to say that $f = g$ a.e.? Or instead do the functions have to be defined on a set of non-zero measure and a.e. refers to some measure $0$ subset?   
I ask because a homework exercise asks me if two real functions are continuous and agree a.e. on a subset of $\mathbb{R}$, are necessarily identically equal. Clearly this is true if the points are not isolated, since if continuous functions disagree at some point, they must disagree on a non-zero measure set since open sets have non-zero measure. Though it need not be if I just use sequences. 
So, what are the requirements to use the phrase a.e.?   
Wolfram definition: A property of $X$ is said to hold almost everywhere if the set of points in $X$ where this property fails is contained in a set that has measure zero. 
This would seem to imply that it is a.e. by default.
 A: To speak of a property holding a.e. on a set $X$ you first need a measure. If $P(x)$ is a property defined for every point $x \in X$, then $P$ is true $\mu$-almost everywhere if $$\mu(\{ x \in X \mid \neg P(x)\}) = 0.$$
Perhaps a bit more precisely, you might require that there is a $\mu$-measurable set $N$ with the property that $\mu(N) = 0$ and $\{ x \in X \mid \neg P(x)\} \subset N$ since the first set is possibly not $\mu$-measurable.
What does it mean for two functions $f,g : \mathbb N \to \mathbb R$ to be equal almost everywhere? Without a measure this question can't be answered. If you specify a measure $\mu$ on $\mathbb N$, they are equal $\mu$-a.e. if $$\mu(\{n \in \mathbb N \mid f(n) \not= g(n)\}) = 0.$$
For instance, if $\mu$ is the counting measure, this states the set of points where $f(n) \not= g(n)$ has zero elements - in this case $f(n) = g(n)$ for all $n$ so that $f$ and $g$ are the same function. On the other hand, if $\mu$ is Lebesgue measure restricted to $\mathbb N$, the fact that $\mu(\mathbb N) = 0$ implies that $\mu(\{n \in \mathbb N \mid f(n) \not= g(n)\}) = 0$ for any two functions $f$ and $g$, so that any two functions $f$ and $g$ are equal a.e.  Different measures yield totally different results.
A: Here's an example to show that your concern about isolated points is justified.
Let $f(x)=0$ for all $x\in{\Bbb R}$ and $g(x) = \max(x,0)$. These two functions are continuous and the are equal at (Lebesgue) a.e. point of $B:=(-\infty,0]\cup\{1\}$. But they are not identically equal on $B$.
A: If you have two functions $f$, $g$ that are defined and continuous on $\mathbb{R}$, then $f=g$ if the two agree a.e. with respect to Lebesgue measure. That may be how the question was meant to be interpreted, because that would be the default for a.e. on $\mathbb{R}$ if no measure is mentioned.
Assuming this interpretation: If the exceptional set is $E$, then you only have to note that $(x-\delta,x+\delta)$ contains a point of $\mathbb{R}\setminus E$ for every $\delta > 0$, which is enough to force $f(x)=g(x)$ for an arbitrary $x$ because $f$ and $g$ are continuous.
