Why is it necessary X and Y beeing bounded when having same moments to be equal

Question

Let $X,Y$ random variables in $[0,1]$ with $E(X^n)=E(Y^n) \,\forall n\in \mathbb N$. I want to show $X\overset{d}{=} Y$.

$$E(e^{itX})=\int\sum \frac{(itx)^k}{k!}dF_X\overset{dom. conv.}{=}\sum\int\frac{(itx)^k}{k!}dF_x=\sum\frac{(it)^k}{k!}\int x^kdF_x\overset{EX^n=EY^n}{=}\sum\frac{(it)^k}{k!}\int x^kdF_Y=\dots=E(e^{itY})$$

I am aware of the fact $EX^n=EY^n$ does not imply $X\overset{d}{=}Y$ in general. I did not use that X and Y are bounded, so what am I missing?

Answer 1

You have to justify interchange of integral and sum. In this case $\int \sum |\frac {(itx)^{k}} {k!}|d_F(x) \leq e^{|t|}$ so we can use Fubini's Theorem.