Measurable functions are not polynomials The problem I have says:

If $f$ is measurable on $\mathbb R$, prove that there is at most a countable number of polynomials $P$ such that $P\overset{\text{a.e.}}{=}f$.

I think I need to show that if $f$ is not a polynomial then it is different almost everywhere from every polynomial. I don't know how to show this though, if it is correct.
Now, if $f$ is a polynomial, then if $f\overset{\text{a.e.}}{=}P_1$ and $f\overset{\text{a.e.}}{=} P_2$, then that means that $P_1\overset{\text{a.e.}}{=}P_2$ where $P_1,P_2$ are polynomials, but doesn't that mean that $P_1=P_2$? I have a hard time finding a counter-example to this. That is where I don't understand the at most countable part, it seems to me that if a function is almost everywhere equal to a polynomial, then it can't be almost everywhere equal to another one. 
Any help would be appreciated!

Since the unique open set with $0$ measure is the empty set, the set where two continuous functions are different is either empty or of positive measure. 
So given a measurable function $f$, there is at most one continuous function $g$ such that $f=g$ almost everywhere. (if $g_1$ and $g_2$ do the job, $g_1=g_2$ almost everywhere, hence (by continuity) everywhere.
