Equal finitely many moments do not determine a law

Put $N=m+1\ge 1$ and $\mu =2^{1-N}{\sum }_{\begin{array}{c}0\le k\le N\\ k\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}\end{array}}\left(\genfrac{}{}{0px}{}{N}{k}\right){\delta }_k,\nu =2^{1-N}{\sum }_{\begin{array}{c}0\le k\le N\\ k\mathrm{o}\mathrm{d}\mathrm{d}\end{array}}\left(\genfrac{}{}{0px}{}{N}{k}\right){\delta }_k\mathrm{.}$ All coefficients are nonnegative real images of the natural binomial coefficients. The binomial theorem at (1,1) and (-1,1) says that the sum of the even and odd coefficient totals is $2^N$, while their difference is zero. Each total is $2^{N-1}>0$. Thus these finite weighted measures are Borel probabilities. They are distinct because $\mu (\{0\})=2^{1-N}>0$ and $\nu (\{0\})=0$.

For a polynomial P define $\mathrm{\Delta }P(x)=P(x+1)-P(x)$. The binomial theorem gives $\mathrm{\Delta }(x^j)={\sum }_{r=0}^{j-1}\left(\genfrac{}{}{0px}{}{j}{r}\right)x^r$ for j>=1, while $\mathrm{\Delta }1=0$. By linearity each application lowers a positive degree by at least one and kills a constant, so ${\mathrm{\Delta }}^{N}P=0$ whenever $\mathrm{deg}P<N$. To compute the iterate, induction gives ${\mathrm{\Delta }}^{r}P(x)={\sum }_{k=0}^r(-1{)}^{r-k}\left(\genfrac{}{}{0px}{}{r}{k}\right)P(x+k)\mathrm{.}$ The base r=0 is P(x). Subtract this expression at x from the expression at x+1: the interior coefficient of P(x+k) is $(-1{)}^{r+1-k}[\left(\genfrac{}{}{0px}{}{r}{k-1}\right)+\left(\genfrac{}{}{0px}{}{r}{k}\right)]=(-1{)}^{r+1-k}\left(\genfrac{}{}{0px}{}{r+1}{k}\right)$, and the coefficients at k=0,r+1 are $(-1{)}^{r+1}$ and one. This proves the induction including both endpoints. Taking r=N, x=0 and P(x)=x^j, j<N, yields ${\sum }_{k=0}^N(-1{)}^k\left(\genfrac{}{}{0px}{}{N}{k}\right)k^j=0$. For j=0 the polynomial is constantly one, so its value at zero is one.

Every required moment is finite because the supports are finite. The difference of the jth moments of the two laws equals $2^{1-N}{\sum }_{k=0}^N(-1{)}^k\left(\genfrac{}{}{0px}{}{N}{k}\right)k^j=0$ for $0\le j\le m=N-1$. This proves the promised failure of determination for every m. When m=0 the witnesses are ${\delta }_0$ and ${\delta }_1$ and only total mass is matched. For m=2 the even law has masses 1/4 at 0 and 3/4 at 2, while the odd law has masses 3/4 at 1 and 1/4 at 3. Their means are both 3/2 and their second moments both 3, yet their masses at zero differ. All finite choices are specified by parity, and no AC is used.