Example of two functions that are equal almost everywhere? 
We shall say that two functions $f$ and $g$ defined on a set $E$ are equal almost everywhere, and write $f(x)=g(x)$ a.e $x\in E$, if the set $\{x\in E: f(x)\neq g(x)\}$ has measure zero.

I just can't wrap my brain around the fact that such functions exist! Certainly, we can take the cantor set which has measure zero but how to pick $f$ and $g$? Is there an example that I am not aware of ? 
 A: There are good answers already, but I think this needs a visual.

A: Consider a measure zero set i.e. $E_1$ such that $\mu(E_1) = 0$. Let $$h(x) = \begin{cases} 0 & x \in E_1^c\\ \text{any value} & x \in E_1 \end{cases}$$
Now consider any function $f(x)$. Then $g(x) = f(x) + h(x)$ satisfies your criteria.
A: If we let $f(x) = \lceil x \rceil$ and $g(x) = \lfloor x \rfloor + 1$ (ceiling and floor functions respectively) then $f$ and $g$ only differ on the set of integers $\mathbb{Z}$, which has measure zero.
A: Consider $\mathbb{R}$ equipped with the Borel $\sigma$-field $\mathcal{B}(\mathbb{R})$ and the Lebesgue measure $\lambda$, i.e. $\lambda$ is the measure on $\mathcal{B}(\mathbb{R})$ which satisfies $\lambda((a,b))=b-a$ for $a,b\in\mathbb{R}$, $a<b$. 
Let $f:\mathbb{R}\to\mathbb{R}$ be any Borel measurable function and suppose we "change" $f$ in at most countably many points. If we call this new function $g$, then $f=g$ almost everywhere (with respect to $\lambda$). In particular, $f$ and $g$ have the same integral over any Borel set.
Let us formalize this: Let $(x_n)_{n\geq 1},(y_n)_{n\geq 1}\subseteq \mathbb{R}$ be two sequences of real numbers and define 
$$
g(x)=
\begin{cases}
y_n\quad &\text{if } x=x_n,\\
f(x)&\text{if }x\neq x_n\text{ for all }n,
\end{cases}
$$
i.e. we have "changed" $f$ along the sequence $(x_n)_{n\geq 1}$. Then
$$
\{x\in\mathbb{R}:f(x)\neq g(x)\}\subseteq \{x_n: n\geq 1\},
$$
which has measure zero.
A: $\newcommand{\R}{\Bbb R}$
There are other sets of measure $0$. For example $Z=\{1\}$ has measure $0$. Consider $f:\R\to \R$ the function defined by
$$f(x)=x,$$
and $g:\R\to\R$ given by $$g(x)=\begin{cases} x &\text{if $x\neq 1$}\\ \pi &\text{if $x=1$}\end{cases}$$
Then $f=g$ a.e. since $f\neq g$ on $Z$.
A: Let $f$ map all real numbers to $0$ and $g$ map all irrationals to $0$ and all rationals to $1$. Then $f$ and $g$ are distinct but equal a.e.
