If $X_{2}$ is a continous random variable then one needs to show that $\nu_A = E(I_A|X_2)$
or alternatively,
$$\nu_A = \operatorname{argmin}\limits_{E(g(X_2) = X_1, E(g(X_2))^2\lneq \infty} E(X_1-g(X_2))^2$$ is a probability on a probability space(A $\in$ Event Space) and hence one needs to verify the three basic properties of a probability on a space.
$1.\nu_{\emptyset}=0$
$2. 0\leq\nu\leq1 \;\;\forall A$
$3.\nu(\uplus A_i)=\sum_i \nu(A_i)$
The first one seems easy to me as it'd follow directly from the properties of indicator function, is it so, or is there some subtle point I'm missing, and I need some help as to how to prove the second and third condition. Would really appreciate some help.
The expected value of an indicator is always a probability since the indicator is a discrete random variable taking on the values $0$ and $1$. So, $$ \begin{array}{lcl} E[I_{A}|X_{2}] &=& 0 \cdot P(I_{A}=0 \,| \,X_{2}) + 1 \cdot P(I_{A}=1 \,| \,X_{2}) = P(A|X_{2})\\ \\ &=& P(I_{A}=1 \, | \, X_{2}). \end{array} $$
So, for example, for disjoint $A_{1}$ and $A_{2}$, $$ E[I_{A_{1} \cup A_{2}}|X_{2}] = P(A_{1} \cup A_{2}|X_{2}) = P(A_{1}|X_{2}) + P(A_{2}|X_{2}). $$