Borel regular measure I stuck with this question, can you help me please.
Is it exist $ \mu$  - Borel regular measure in $[0,1]$ so that to all polynomial $p$ one has:
$\int_{[0,1]}p(t)d \mu(t)=p'(0)$?
Thanks a lot!
 A: For such a measure $\mu$ we have, for $n=0$ and $n\geq2$,
$$\int_{[0,1]}t^n d \mu(t)=0$$ and hence 
 $\int_{[0,1]}p(t) d \mu(t)=0$ for all polynomials without linear term. 
The sequence 
$p_n(t)=1-\sum_{j=1}^n \left|{1/2 \choose j}\right| (1-t^2)^j$ converges to $t$ uniformly on $[0,1]$, which implies 
$$0=\int _{[0,1]}p_n(t)d \mu(t)\to\int _{[0,1]} t\,d \mu(t)=1; $$ 
 a contradiction.
A: By Riesz theorem there exist isometric isomorphism 
$$
I:M([0,1])\to C([0,1])^* :\mu\mapsto\left(x\mapsto \int\limits_{[0,1]}x(t)d\mu(t)\right)
$$
between Borel $\sigma$-additive measures and bounded functionals on $C([0,1])$. You can check that linear functional defined on dense subspace consisting of polynomials 
$$
\hat{f}:P([0,1])\to \mathbb{C}: p\mapsto p'(0)
$$
is not bounded. Contradiction.
A: We can use Bernstein polynomials: if $f$ is continuous on $[0,1]$, its Bernstein polynomial of degree $n$ is defined as 
$$P_n(x):=\sum_{k=0}^n\binom nkx^k(1-x)^{n-k}f\left(\frac kn\right).$$
We can see that 
$$P'_n(0)=\frac{f\left(\frac 1n\right)-f(0)}n$$
and that $P_n$ converges uniformly to $f$ on $[0,1]$. Hence, if we assume $\mu$ finite, we should have that 
$$\int_{[0,1]}fd\mu=\lim_{n\to +\infty}\frac{f\left(\frac 1n\right)-f(0)}n.$$
But this limit doesn't need to exist, as the function 
$$
f(x)=\begin{cases}x\sin\left(\frac 1x\right)&\mbox{ if }x\neq 0,\\
0&\mbox{ if }x=0.
\end{cases}$$
