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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?

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up vote 4 down vote accepted

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.

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+1 for the example, -1 for the awkward notation. Seriously, why don't you write: The characteristic function $\chi_{C}$ of the Cantor set is of Baire class $1$ because it is the pointwise limit of the sequence of continuous functions given by$g_{n}(x) = \max{\{0, 1 - n d(x,C)\}}$. Or even easier: The pre-image of an open set under $\chi_{C}$ is either $\emptyset$, $[0,1]$, $C$ or $[0,1] \smallsetminus C$, each of which is an $F_{\sigma}$, hence $\chi_{C}$ is of Baire class $1$. –  Theo Buehler Jan 12 '11 at 23:30
    
Thanks! (Much easier than I feared.) –  Bruce Jan 12 '11 at 23:35
    
@Theo: Because in answering the main question, I wanted more rigour. –  Ricky Demer Jan 12 '11 at 23:46
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