Product measure and independence What's the difference between product measure and independence in probability?
$$
(\mu_1\times\mu_2)(B_1\times B_2)= \mu_1(B_1)\mu_2(B_2)
$$
$$
P(A_m\cap A_k)= P(A_m)P(A_k)
$$
Are product measure it self independent?
 A: There is an equivalence in the following sense.

*

*Suppose $(\Omega,\mathscr{F},\mathbb{P})$ is a probability space.

*Let $X=\{X_j:j\in J\}$ be a collection of random variables taking values in spaces $(S_j,\mathscr{S}_j)$.

*For each $j$, let $\mu_j=\mathbb{P}\circ X^{-1}_j$ be the law of $X_j$, that is $\mu_j(B)=\mathbb{P}[X_j\in B]$ for any $B\in\mathscr{S}_j$.


*

*If $J$ is finite then: $\{X_j:j\in J\}$ is independent (e.i, for any $J'\subset J$ and sets $B_j\in\mathscr{S_j}$, $j\in J'$,
$$\mathbb{P}\Big[\bigcap_{j\in J'}\{X_j\in B_j\}\Big]=\prod_{j\in J'}\mathbb{P}[X_j\in B_j]\tag{1}\label{one}$$
) if and only if the (joint) law of $X=\{X_j:j\in J\}$ is the product measure $\bigotimes_{j\in J}\mu_j$.


*If $J$ is an arbitrary set (finite or infinite), and each space $(S_j,\mathscr{S}_j)$ is nice (a separable metric space with its Borel $\sigma$--algebra for instance; $\mathbb{R}$ with its Borel $\sigma$--algebra for example), then $\{X_j:j\in J\}$ is independent (i.e. $\eqref{one}$ holds for ay finite collection $J'\subset J$) then the law of $X=\{X_j:j\in J\}$ is the product measure $\bigotimes_{j\in J}\mu_j$ .

The proof of (2) follows from Kolmogorov's extension theorem. The "nice" condition is needed in order to appeal to measure theoretic results that guarantee existence of a unique measure in an arbitrary product of  probability spaces that satisfy a projection property.
Leo Breiman's on Probability is an excellent source where the details are explained in a very elegant way.
A: Independence of two random variables means the joint measure induced by them both can be factorized. At a very basic heruistic level, if you have random variable $X$, then it introduces a probability density function $f_{X}$ on the sample space $D$. We need to assume $f_{X}$ is measurable (so it work nicely) and there exists an associated measure $\mu_{X}$ such that
$$
\int_{D}h(x)f_{X}dx=\int_{D}g(x)(\mu_{X} dx), \forall h\in L^{1}(D)
$$
Now the independence of $X,Y$, as two random variables just means the induced measure of $Z=(X,Y)$ on $D\times D$ factorizes:
$$
\mu_{(X,Y)}=\mu_{X}*\mu_{Y}
$$
Without the measure theory jargon, this means that we have
$$
P(X\in E_{i},Y\in E_{j})=P(X\in E_{i})*P(Y\in E_{j}), E_{i}, E_{j}\subset D
$$
where $E_{i},E_{j}$ are any measurable (nicely behaved) subsets of $D$. 
You can find a lot more expert input from this blogpost written by a real master. 
