Independence is factorization of joint probabilities. Measure theory makes it possible to state independence for events, sigma-fields, random variables, and stochastic processes using the same product principle.
Independence of sigma-fields
The cleanest definition is made at the level of information.
Definition. \(\sigma\)-fields \(\mathcal G_i\subseteq\mathcal F\) are independent if for every finite set of indices and every \(G_i\in\mathcal G_i\), \[ P\left(\bigcap_iG_i\right)=\prod_iP(G_i). \] Random variables are independent when their generated \(\sigma\)-fields are independent.
Proposition. If \(X_1,\ldots,X_n\) are independent and \(f_i\) are measurable, then \(f_1(X_1),\ldots,f_n(X_n)\) are independent. If the product is integrable, then \[ \mathbb E\prod_{i=1}^nf_i(X_i) =\prod_{i=1}^n\mathbb E f_i(X_i). \]
Proof. Since \[ \sigma(f_i(X_i))\subset\sigma(X_i), \] the event-factorization definition immediately preserves independence. For nonnegative \(f_i\), the expectation factorization follows first for indicators, then simple functions, then measurable functions by monotone convergence. Integrable signed products follow by positive and negative parts under the stated integrability.
Checking on generators
Independence does not need to be checked on every measurable set.
Theorem. If \(\mathcal C_i\) are pi-systems with \(\sigma(\mathcal C_i)=\mathcal G_i\), and the factorization identity holds for all finite choices \(C_i\in\mathcal C_i\), then \(\mathcal G_i\) are independent.
Proof. For two classes, fix \(C_1\) and collect all \(G_2\) for which \(P(C_1\cap G_2)=P(C_1)P(G_2)\). This is a lambda-system containing \(\mathcal C_2\), hence contains \(\mathcal G_2\). Then fix \(G_2\) and repeat in the first coordinate. The finite case follows by induction.
Products and Fubini-Tonelli
Independence turns joint laws into product measures and makes iterated integration valid.
Proposition. If \(X\) and \(Y\) are independent with laws \(\mu\) and \(\nu\), then the joint law of \((X,Y)\) is \(\mu\otimes\nu\). Consequently, for nonnegative or integrable \(g\), \[ \mathbb E[g(X,Y)]=\int\!\!\int g(x,y)\,\mu(\mathrm{d}x)\nu(\mathrm{d}y). \]
Proof. The rectangle identity follows from independence: \(P((X,Y)\in A\times B)=P(X\in A)P(Y\in B)=\mu(A)\nu(B)\). Rectangles generate the product \(\sigma\)-field, so uniqueness of product measures gives the joint law. The integral identity is Tonelli's theorem for nonnegative \(g\) and Fubini's theorem for integrable \(g\).
Example (pairwise but not mutual independence). Let \(U,V\) be independent fair bits and define addition modulo \(2\) by \[ X=U,\qquad Y=V,\qquad Z=U\oplus V. \] Each pair is independent: for example, fixing \(U=u\), the map \(V\mapsto U\oplus V\) remains a fair bit. But \[ X\oplus Y\oplus Z=0 \] surely, and \[ P(X=0,Y=0,Z=0)=\frac14\ne\frac18 =P(X=0)P(Y=0)P(Z=0). \] Pairwise factorization does not imply mutual factorization.
Proposition (independence from a joint density). If \((X,Y)\) has joint density \(p_{X,Y}\) and marginal densities \(p_X,p_Y\), then \[ X\perp Y \quad\Longleftrightarrow\quad p_{X,Y}(x,y)=p_X(x)p_Y(y) \quad\text{for Lebesgue-a.e. }(x,y). \]
Proof. The factorized density integrates over every rectangle \(A\times B\) to \(P_X(A)P_Y(B)\), so rectangle uniqueness gives independence. Conversely, independence says the joint law is \(P_X\otimes P_Y\); its Radon-Nikodym derivative with respect to product Lebesgue measure is \(p_Xp_Y\). Uniqueness of densities gives the a.e. identity.