Does E[f(X)g(Y)] = E[f(X)] E[g(Y)] for bounded, continuous f & g imply independence? I understand that independence of random variables $X$ and $Y$ implies that $E[f(X)g(Y)] = E[f(X)]*E[g(Y)]$ for measurable $f$ and $g$, because independence of $X$ and $Y$ means that their $\sigma$-algebras are independent, and $\sigma(f(X)) \subset \sigma(X), \sigma(f(Y)) \subset \sigma(Y)$.
However,  I'm having trouble proving that if $E[f(X)g(Y)] = E[f(X)]*E[g(Y)]$ for all bounded and continuous $f$, $g$, then $X$ and $Y$ are independent.  
This is in my notes for a class, but I'm having some difficulty proving it (not HW; studying for an exam next week).
My notes also say "by DCT" (dominated convergence theorem, I assume), but I can't figure out how to use DCT here, since my understanding of DCT is that it goes the other way, from convergence of integrands to convergence of integrals (i.e. if $f_n \rightarrow f$, and $|f|<g, g$ integrable, then $\int f_n d\mu\rightarrow \int f d\mu$).
Can someone help me out?
 A: Given $a<b$ it is easy to construct a sequence of continuous functions $(f_n)$ such that $0\leq f_n \leq 1$ and $f_n(x) \to 1$ for $x \in [a,b]$, $f_n(x) \to 0$ for $x \notin [a,b]$. Similarly take $c<d$ and choose continuous functions $(g_n)$ such that $0\leq g_n \leq 1$ and $g_n(x) \to 1$ for $x \in [c,d]$, $g_n(x) \to 0$ for $x \notin [c,d]$. Then we can apply DCT in $Ef_n(X)g_n(Y)=Ef_n(X)Eg_n(Y)$ to get $P(X\in [a,b],Y \in [c,d]) =P(X \in [a,b])P(Y\in [c,d])$. This implies that $X$ and $Y$ are independent. 
Explicit definition of $f_n$: $f_n(x)=1$ if $a\leq x \leq b$, $0$ if $x <a-\frac 1 n$ or $x >b+\frac 1 n$, $n(x-a+\frac 1 n)$ if $a-\frac 1 n \leq x \leq a$ and $n(b+\frac  1n -x)$ if $b \leq x \leq b+\frac 1 n$.
A: Without being rigorous,
\begin{align}
\mathbb{P}(X\in A, Y\in B)&=\mathbb{E}(I_A(X)I_B(Y))\quad\quad\mbox{ by definition}\\
&=\mathbb{E}(I_A(X))\mathbb{E}(I_B(Y))\quad\quad\mbox{ by hypothesis}\\
&=\mathbb{P}(X\in A)\mathbb{P}( Y\in B)\quad\quad\mbox{ by definition}
\end{align}
