Baire class 1 functions for which $\int_0^1 fp=0$ for all polynomials $p$ Suppose that $f$ is a class 1 Baire function defined on ${}[0,1]$ such that $\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$. 
Of course, $f$ must be 0 almost everywhere. Can we conclude that in fact $f$ is zero except at countably many points? If yes, how high up in the Baire hierarchy can we go and still be able to reach this conclusion? If no, what if all discontinuities of $f$ are jump discontinuities: Are there non-trivial examples in this case, other than a continuous function that we redefined at countably many points? 
 A: No.

Define $g : \mathbb{N} \times [0,1] \to \mathbb{R}$ by $f(n,x) := \operatorname{max}(0,1+((-n)\cdot \operatorname{inf}(\{|x+(-y)| : y\in (\operatorname{Cantor} \operatorname{set})\}))).$
$g$ is continuous, so it is in Baire class 0.
Define $f : [0,1] \to [0,1]$ by $f(x) := \displaystyle\lim_{n\to \infty} g(n,x)$.
$f$ is in Baire class 1, and is the characteristic function of the Cantor set.
$f$ is 0 almost everywhere, but not at all but countably many points.
$\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$.
Therefore there exists a Baire class 1 function $f$ such that $\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$ and it is not the case that $f$ is zero at all but countably many points.
QED

If all discontinuities of $f$ are jump discontinuities, then the set of discontinuities of $f$ is countable, see http://en.wikipedia.org/wiki/Regulated_function.  If $f$ additionally satisfies you integral condition, then $f$ is 0 almost everywhere, so in particular $f$ is 0 densely often, in which case $f$ is 0 everywhere it is continuous.  Therefore, if $f : [0,1] \to \mathbb{R}$ satisfies $\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$ and all discontinuities of $f$ are jump discontinuities, then $f$ is 0 at all but countably many points, and so in particular $f$ is the 0 function redefined at countably many points.

QED
Note that $f$ is not assumed to be in Baire class 1.
