Formal vs. Intuitive independence of random variables Here is an issue from probability theory which I feel is being neglected a lot, and is bothering me.
The formal definition of independent random variables is
$$f_{X,Y}(x,y) = f_X(x)f_Y(y)$$
Where $f_X$ etc. are the probability densities, and $X,Y$ are Borel-measurable functions on the probability space. That (I feel) is a very strong thing to demand. For example, demanding this is stronger than demanding that 
$$ E(XY) = E(X)E(Y) => \int_{\Omega}XYdP = \int_{\Omega}XdP\int_{\Omega}YdP$$
i.e. that integral of product equals product of the integral, which, if you take any two "random" (here the term is used informally) functions, will simply not be true.
On the other hand, the verbal, intuitive definition of independence: "knowledge of the result of one variable contributes nothing to the knowledge of the other" is extremely basic. In almost all instances the question of independence of random variables is clear - it is "obvious", for instance, that if we toss one coin, and do some simple operation with the result, and then toss another coin, and do the same operation, the results will be independent.
So, is there an explanation to such a big gap between formal and intuitive definitions? In most cases I know of, when we give a mathematical definition to an 'intuitive' concept, the mathematical definition is usually simple as well. As it is, the independence is implied in almost all probability/statistical questions I have seen to date, and never have I seen the need to prove it... simply due to the 'obviousness' of the answer and the complexity of the proof.
Is there a theorem which simply proves the independence in those simple, 'intuitive' cases?
 A: Consider what happens if, for example, $f_{XY}(x,y) > f_X(x) f_Y(y)$ 
for all points $(x,y)$ in the rectangular region $[x_1,x_2]\times[y_1,y_2]$ (that is, all
points such that $x_1 \le x \le x_2$ and $y_1 \le y \le y_2$).
Now consider the events $A$ and $B$, where $A$ is the event that $x_1 \le X \le x_2$
and $B$ is the event that $y_1 \le Y \le y_2$.
Supposing the probability of $B$ is not zero, 
what is the probability of $A$ given that $B$ occurs?
Writing this out as a conditional probability,
$$P(A | B) = \frac{P(A \cap B)}{P(B)}.$$
We also have
$$P(A \cap B) = \int_{y_1}^{y_2} \int_{x_1}^{x_2} f_{XY}(x,y) \, dx \,dy$$
and
\begin{eqnarray}
P(A)P(B) &=& \left( \int_{x_1}^{x_2} f_X(x) \, dx \right)
 \left(\int_{y_1}^{y_2} f_Y(y) \,dy  \right)\\
&=& \int_{y_1}^{y_2} \int_{x_1}^{x_2} f_X(x) f_Y(y) \, dx \,dy.
\end{eqnarray}
But due to our initial assumption that $f_{XY}(x,y) > f_X(x) f_Y(y)$
over this region,
$$\int_{y_1}^{y_2} \int_{x_1}^{x_2} f_{XY}(x,y) \, dx \,dy
> \int_{y_1}^{y_2} \int_{x_1}^{x_2} f_X(x) f_Y(y)\, dx \,dy.$$
Therefore $P(A \cap B) > P(A)P(B)$, which means that
$$P(A | B) > \frac{P(A)P(B)}{P(B)} = P(A).$$
So indeed, knowing that $y_1 \le Y \le y_2$ would tell us something about the
probability that $x_1 \le X \le x_2$ in the case of this joint probability distribution.
These two random variables do not satisfy our intuitive definition of independence.
It would of course take more work than this to show that the usual formal definition of
independence is really equivalent to the intuitive definition, but at least this should give a hint as to why such a strong definition is necessary.
A: $E(XY) = E(X)E(Y)$ is clearly insufficient for independence: consider $X$ being any continuous random variable with expected value of $0$ and $Z$ being $\pm1$ independently with equal probability.  If $Y=XZ$ then the expectation of the product is the product of the expectations, but since $|X|=|Y|$ they are clearly not independent.
So you need more.  If your intuition leads you to consider $E(g(X)h(Y)) = E(g(X))E(h(Y))$ for all functions $g$ and $h$ sufficiently well-behaved to make both sides finite as a reasonable property of independence, then your intuition comes closer to the formal definition. 
