If $g(x) \ne f(x)$ almost everywhere, then polynomial transformation of $P(g(x)) \ne P(f(x))$? Suppose that we have two functions $f(x)$ and $g(x)$. We know that $f(x) \ne g(x)$ on the interval, except for measure zero subsets of $[a,b]$. Assume that both functions are positive everywhere and both are strictly increasing almost everywhere (i.e. there is a subset $S$ of the interval $[a,b]$, where $[a,b] \setminus S$ has measure $0$, and $f, g$ are strictly increasing on $S$).
Under these conditions on $f$ and $g$ (positive, strictly increasing a.e.), is it possible that there exists a non-constant polynomial transformation $P$ such that $P(f(x)) = P(g(x))$ almost everywhere on $[a,b]$?
 A: I think that the following is a counterexample:
Let $f(x)=\sin(x), g(x)=-\cos(x)$ on $[0, \frac{\pi}{2}]$ and $P(X)=X^2(1-X^2)$.
Then 
$$P(f(x))=P(g(x))=\sin^2(x) \cos^2(x)$$
A: It is possible to have positive, strictly increasing functions $f, g: [a,b] \to \mathbb{R}$ such that $f(x) \ne g(x)$ for all $x \in [a,b]$, yet there is a non-constant polynomial $P$ such that $P(f(x)) = P(g(x))$ for all $x \in [a,b]$.
To find an example, let's start with the polynomial $P$: take $P(X) = (X - 2)^3 - (X - 2)$.
Then, plot the points $(X,Y)$ where $P(X) = P(Y)$. We get:

The key is to look for a point on the graph where there is a tangent line with positive slope in the first quadrant. Above, we see this occurs at the point $(3,1)$. So we can set $(f(t), g(t))$ to be a parametrization of the curve starting from this point and moving in the positive $x$- and $y$-directions. We note that by construction, $f$ and $g$ are increasing and positive, and they satisfy $P(f(t)) = P(g(t))$.
Explicitly, if we parametrize the above curve we get the following example, which also uses trigonometry like N. S.'s nice answer:
\begin{align*}
P(X) &= (x-2)^3 - (x-2) \\
f(x) &= 2 + \frac{1}{\sqrt{3}} \cos x + \sin x \\
g(x) &= 2 + \frac{1}{\sqrt{3}} \cos x - \sin x
\end{align*}
These are always positive and are both strictly increasing on some intervals, e.g. on the interval $[-2\pi/3,-\pi/3]$.
The algebra is a bit messy, but you can verify that $P(f(x)) = Q(f(x))$ with Wolfram Alpha.
In general, for any polynomial $P$ we can plot the set of $(X,Y)$ coordinates where $P(X) = P(Y)$, and anywhere where the tangent curve has positive slope we can read off a pair of functions $f, g$ satisfying $P(f(t)) = P(g(t))$.

Of course, there are also examples where no such polynomial exists. For instance, take
\begin{align*}
f(x) &= x^4 \\
g(x) &= x^2
\end{align*}
on the interval $[1,2]$.

Note that for either example, you can make the function increasing on all of $\mathbb{R}$ and not just $[a,b]$, by extending $f$ appropriately to values $x < a$ and $x > b$. 
