If $\mathbb E[X^nY^m]=\mathbb E[X^n]\mathbb E[Y^m]$ for all $n,m$, then $X$ and $Y$ are independent. I had an exam this morning, and I had to prove that for $X$ and $Y$ bounded, if for all $n$ and all $m$, $$\mathbb E[X^nY^m]=\mathbb E[X^n]\mathbb E[Y^m],$$then $X$ and $Y$ are independents. Using Characteristic function I think that I could justify it using the fact that $$\cos(tX)=\sum_{k=0}^\infty \frac{(-1)^kt^{2k}X^{2k}}{(2k!)}\quad \text{and}\quad \sin(tX)=\sum_{k=0}^\infty \frac{(-1)^kt^{2k+1}X^{2k+1}}{(2k+1)!},$$
but how can I do without characteristic function, and out measure theory ? (it's a lecture of introduction of probability, so we don't have tools of measure theory as DCT or MCT, neither characteristic function).

My definition of the expectation is : for $X$ s.t. $\mathbb E|X|<\infty $, $$\mathbb E[X]=\lim_{n\to \infty }\mathbb E[X_n]=\sum_{k\in\mathbb Z}\frac{k}{2^n}\mathbb P\left(X_n=\frac{k}{2^n}\right),$$
where $X_n=2^{-n}\lfloor 2^nX\rfloor$.
 A: It would be interesting to known if there is a short proof of this.

Here is a (not so short) one...
Let $X,Y$ be bounded random variables.  We start with
$$
\mathbb E[X^nY^m] = \mathbb E[X^n]\mathbb E[Y^m]\qquad \text{for all }n,m \in \mathbb N
\tag1$$
Using linear combinations, we deduce
$$
\mathbb E[f(X)g(Y)] = \mathbb E[f(X)]\mathbb E[g(Y)]\qquad \text{for all polynomials }f,g
\tag2$$
Using Weierstrass approximation (uniform convergence) we deduce
$$
\mathbb E[f(X)g(Y)] = \mathbb E[f(X)]\mathbb E[g(Y)]\qquad \text{for all continuous functions }f,g
\tag3$$
Using dominated convergence (pointwise a.e. convergence) we deduce
$$
\mathbb E[f(X)g(Y)] = \mathbb E[f(X)]\mathbb E[g(Y)]\qquad \text{for all bounded measurable functions }f,g
\tag4$$
In particular, $(4)$ holds for all indicator functions.  That is the definition of "$X$ and $Y$ are independent".
A: Are moment generating functions OK? Since $e^{sX}=\sum_{n=0}^\infty\frac{s^nX^n}{n!}$ and $e^{tY}=\sum_{m=0}^\infty\frac{t^m Y^m}{m!}$, by linearity
$$M_{X,Y}(s,t)=\mathbb E\left[e^{sX}\cdot e^{tY}\right]=\mathbb E\left[e^{sX}\right]\cdot\mathbb E\left[e^{tY}\right]=M_X(s)\cdot M_Y(t).$$
So $X,Y$ are independent.
This is probably similar to whatever method you used with characteristic functions, but I think normally MGFs are regarded as more elementary.
