Does there exist a function $p:\mathbb{Q}\cap[0,1]\to\mathbb{R}$ st. for any continuous $f:[0,1]\to\mathbb{R}$ we have $\sum_{x\in \mathbb{Q}\cap [0,1]} f(x)p(x) =\int_0^1 f(x)dx$?
If this is not possible, then how about other classes of functions for f, for example, infinitely differentiable functions, analytic functions, polynomials.
Interpreting the statement in the sense that the sum over the rationals on the left hand side converges absolutely, then no, it is not possible.
Let $r_n$ be an enumeration of $\mathbb Q\cap [0,1]$ and let $p_n:=p(r_n)$.
Then $\sum_n p_n = 1$ (convergence is absolute). Let $\epsilon>0$ be given and let $N$ be such that $\sum_{n\geq N}|p_n|<\epsilon$. Let $f$ be a continuous (or even $\mathcal C^{\infty}$) function $[0,1]\to[0,1]$ such that $f(r_n)=1$ for all $n<N$ and $\int_0^1 f(x) dx<\epsilon$. Then, $$ \int_0^1f(x)dx -\sum_{n\geq N} f(r_n)p_n=\sum_{n<N}p_n>1-\epsilon. $$ On the other hand, $$ \left|\int_0^1f(x)dx -\sum_{n\geq N} f(r_n)p_n\right|\leq 2\epsilon $$ and if we start this argument with $\epsilon<1/3$ we reach a contradiction.
