A question related to independence and expectation I am trying to prove the following Lemma:

Random variables $\{X_i\}_{i=1}^\infty$ are independent iff for all
  measurable functions $\phi_1, \ldots, \phi_n: \mathbb{R}^d \to
\mathbb{R}_+$,
$$\mathbb{E}\left[ \prod_{j=1}^n \phi_j(X_j) \right] = \prod_{j=1}^n
\mathbb{E}\left[ \phi_j (X_j)\right]$$

In the proof part, the book (p. 68) says that it can be proved by using the "standard machine", i.e. using the elementary functions, then simple functions then the MCT (by taking limits?). As I am not good at using this machine, can anyone show a full (with details) proof of this assertion?
The second thing I confused when I tried to use standard machine is that, here $X_j$ are random variables hence they are measurable functions too (in addition to $\phi$). I wonder if this needs extra effort in proof.
 A: This is not a complete, detailed proof, but maybe I can say a little more about what you need to do.
So I'm assuming the theorem is clear if we replace "all measurable functions $\phi_1,\ldots,\phi_n$" by "all elementary functions $\chi_E$ (where $\chi_E(x) = 1$ when $x\in E$ and 0 otherwise)."  So you really are just curious about proving the left->right implication for more general functions than elementary ones.
First, if you show the implication for n=2, then the general $n$ case is not really any different, so let's just take n=2.
So to start, assume the $X_i$ are independent.  We have the above equality for any two elementary functions $\chi_E$, $\chi_F$.  To show the equality also holds for simple functions, first note that any simple function can be written as a finite linear combination of elementary functions.  With this in mind, note that
$\mathbb{E}[\phi_1(X_1)\cdot \phi_2(X_2)] = \mathbb{E}[\phi_1(X_1)]\mathbb{E}[\phi_2(X_2)]$
implies (by lineary of expection) that
$\mathbb{E}[\alpha\phi_1(X_1)\cdot \phi_2(X_2)] = \mathbb{E}[\alpha\phi_1(X_1)]\mathbb{E}[\phi_2(X_2)]$
and also that both
$\mathbb{E}[\phi_1(X_1)\cdot \phi_2(X_2)] = \mathbb{E}[\phi_1(X_1)]\mathbb{E}[\phi_2(X_2)]$ 
and 
$\mathbb{E}[\phi_3(X_1)\cdot \phi_2(X_2)] = \mathbb{E}[\phi_3(X_1)]\mathbb{E}[\phi_2(X_2)]$
imply by linearity of expectation
$\mathbb{E}[(\phi_1+\phi_3)(X_1)\cdot \phi_2(X_2)] = \mathbb{E}[(\phi_1+\phi_3)(X_1)]\mathbb{E}[\phi_2(X_2)]$.
These facts are all you need to go from elementary to simple functions.
Next, recall that any nonnegative integrable function is the limit of an increasing sequence of simple functions.  So note that if already we know
$\mathbb{E}[\phi_1(X_1)\cdot \phi_i(X_2)] = \mathbb{E}[\phi_1(X_1)]\mathbb{E}[\phi_i(X_2)]$
for $\phi_i$, $i=2,3,\ldots$, an increasing sequence of functions then by the monotone convergence theorem,
$\lim \mathbb{E}[\phi_1(X_1)\cdot \phi_i(X_2)] = \mathbb{E}[\phi_1(X_1)\cdot \lim(\phi_i(X_2))]$
and 
$\lim \mathbb{E}[\phi_1(X_1)]\mathbb{E}[\phi_i(X_2)] = \mathbb{E}[\phi_1(X_1)]\mathbb{E}[\lim(\phi_i(X_2))]$
Since the terms in the limit of the left sides of these above two equations are equal, so are the right sides.
Everything you need I believe is all here (I'm actually thinking I may have given something pretty close to a full proof, just see that this works for general n), and you don't need to think too much about measurability (all those considerations should be within statements like "all nonnegative integrable functions are the sup of simple functions").
A: Depending on what you know or on how independence is defined, you might not need standard machine. Suppose


*

*$X,Y$ are independent iff $E[f(X)g(Y)]=E[f(X)]E[g(Y)]$

*$X_1,X_2,...,X_n$ are independent iff $E[\prod f_i(X_i)]=\prod E[f_i(X_i)]$

*$X_1,X_2,...,X_n$ are independent iff any subset is independent.

*$X_1,X_2,...,$ are independent iff any finite subset is independent.

*$X_1,X_2,...,$ are independent iff for any n, $X_1,X_2,...,X_n$ is independent.


Then (2) and (5) give you your answer.
As for proving (2), you get that with strong induction, (1) and (3).
As for proving (1), (only if) is obvious. For (if), let $A,B \in \mathscr B$, choose
$$f(x) = 1_{x \in A}, g(y) = 1_{y \in B}$$
I think. I'm extending the proof here for discrete from $x=a$ to $x \in A$.
