Dirac Mass and Joint Distribution. Take two spaces $X_1$ $X_2$ and consider $\mu\in\mathcal{P}(X_1\times X_2)$ i.e the space of Borel probability measures. Let $\mu_1,\mu_2$ be Borel probability measures on $X_1,X_2$, with $\mu_1$ a dirac mass.
Is it true that if $\mu$ has marginals $\mu_1,\mu_2$ then $\mu=\mu_1\times \mu_2$ i.e the product measure? If so how do I prove this result?
 A: Writing this up for completeness, even though the question is very old:
Let $\mu_1 = \delta_x$. Let $A \in \mathcal{B}(X_1)$, $B \in \mathcal{B}(X_2)$. We want to show that
$$
\mu (A \times B) = 
\begin{cases}
   \mu_2 (B) &\text{if}\:\: x \in A \\
   0 &\text{if}\:\: x \notin A.
\end{cases}
$$
If $x \notin A$, then $\mu (A\times B) \leq \mu (A \times X_2) = \mu_1 (A) = 0$ which shows the second case.
Then, for $x \in A$, this implies $\mu(A^{c} \times B) = 0$. Hence
$$
\mu (A \times B) = \mu (X_1 \times B) - \mu(A^c \times B) = \mu_2(B)
$$
completing the proof.
A: Since $\mu_i$ is a Dirac mass for each $i = 1, 2$, we can find $x_i \in X_i$ such that $\mu_i = \delta_{x_i}$. Now if we write
$$ E_1 = (X_1\setminus\{x_1\})\times X_2, \qquad E_2 = X_1 \times (X_2\setminus\{x_2\}), $$
then we have
$$\mu(E_1) = \mu_1(X_1\setminus\{x_1\}) = 0 \qquad\text{and}\qquad \mu(E_2) = \mu_2(X_2\setminus\{x_2\}) = 0. $$
So it follows that
$$ \mu((X_1\times X_2)\setminus\{(x_1,x_2)\}) = \mu(E_1 \cup E_2) = 0. $$
This tells that
$$ \mu = \delta_{(x_1,x_2)} = \delta_{x_1} \otimes \delta_{x_2}. $$
