 Solution

Let $X\sim \mathrm{Bin}(n,t)$ and $Y\sim \mathrm{Bin}(n,s)$ be independent. The displayed Bernstein polynomial isGiven $\epsilon >0$, uniform continuity of $f$ supplies $\delta >0$ such that changing both coordinates by less than $\delta$ changes $f$ by less than $\epsilon$. If $M=\Vert f{\Vert }_{\infty }$, then Chebyshev inequality and
$\mathrm{Var}(X/n),\mathrm{Var}(Y/n)\le 1/(4n)$ give, uniformly in $(t,s)$,Thus $B_{n}f\to f$ uniformly.

Each $B_{n}f$ is a polynomial, whose iterated integrals may be interchanged term by term. Uniform convergence permits passage to the limit in both iterated integrals, proving the asserted continuous-function form of Fubini's theorem.

Now suppose every monomial moment of $f$ vanishes. By linearity,
$\int \int pf=0$ for every two-variable polynomial $p$. The just-proved Bernstein approximation gives polynomials $p_n\to f$ uniformly. Thereforeso $\boxed{f=0}$. Statement (i) is true.