Why does this statement not hold when $m(E)=0$? 
"Suppose $f$ and $g$ are continuous on the measurable set $E$. If $f = g$ a.e. on $E$, then $f=g$ on $E$."

I have shown that this statement is true when $E = [a,b]$ and I have shown that the statement is true for $E$ when $m(E)\neq 0$. However, my proof for the latter case never explicitly used the fact that $m(E)\neq 0$.
I have been told that the statement does not hold when $m(E)=0$. However, I am unconvinced (seems to me it should be vacuously true) and unable to prove it.
Could somebody please share with me a quick, not too complicated proof for why the statement is not true when $m(E) = 0$?
Thank you. 
 A: *

I have shown that this statement is true when $E=[a,b]$  and I have
  shown that the statement is true for $E$ when $m(E) \not=0$.

*
Well done on the former.  As for the latter ... not so good.
The principle at work here has little to do with measure theory.  It is simply this: if $f$ and $g$ are continuous functions that agree on a dense subset of $E$ then they agree on all of $E$.
So if $N\subset [a,b]$ is a set of measure zero then you can use  the fact that $[a,b]\setminus N$ is a dense subset of $[a,b]$.
If, however, $N\subset E$ is a set of measure zero it does not follow that $E\setminus N$ is a dense subset of $E$.
For example take $E=[0,1] \cup \{2\}$ and $N = \{2\}$.  $N$ is a measure zero set but you simply cannot conclude that two continuous functions that agree on $[0,1]$ would have to agree on $E$.
A: Statement:$$f=g\text{ a.e. on }E$$ is the same statement as: $$f1_E=g1_E\text{ a.e.}$$
If $m(E)=0$ then this statement is true for any pair of continuous functions $f,g$.
But $m(E)=0$ does not imply that $E=\varnothing$.
If  $f(x)\neq g(x)$ for some $x\in E$ then the statement: $$f=g\text{ on }E$$ is false.
A: The assertion is false for some $E$ with $m(E)> 0.$ For example it fails for $E=[0,1]\cup \{2\}.$ The assertion is false for any $E$ with $m(E)=0.$ Here just take $f=0,g=1$ on $E.$
A: Let $E = \{\ast\}$ be some set with a single element.
Equip it with the trivial sigma-algebra and a measure $m$ that is constant $0$.
Let $f: E \to \{1,2\}$ be the constant $1$ function.
Let $g: E \to \{1, 2\}$ be the constant $2$ function.
Trivially, $f$ and $g$ are continuous. Trivially, it holds $f=g$ almost everywhere (the only point $\ast$ does not count, because $m(\{\ast\}) = 0$). However, they are obviously not equal.

If $E$ has to be a subset of $\mathbb{R}$ equipped with the trace-sigma algebra of the standard Borel sigma algebra of $\mathbb{R}$, the argument still holds: simply take $E=\{0\}$, or pick any other real number.
