Product measure with a Dirac delta marginal Let $(S,\mathcal F)$ be a measurable space, and let $\nu \in\mathcal P(S,\mathcal F)$ be a probability measure on $(S,\mathcal F)$. Fix some $x\in S$ and consider Dirac measure $\delta_x$. Would like to prove 
If $\mu \in \mathcal P(S×S,\mathcal F\otimes \mathcal F)$ and has marginals $ν$ and $δ_x$  $then$ $μ=ν×δ_x$
So we should show $μ(A×B)=ν(A)δ_x(B)$ for $∀A,B∈\mathcal F$. 
If $x∉B$ then right-hand side is $0$ but so is the left-hand side since $μ(A×B)≤μ(S×B)=δ_x(B)=0$.
How to deal with the case $x∈B$ ?
 A: If $x\in B$, then for each $A\in\mathcal F$, $$\nu(A)=\mu(A\times S)\geqslant \mu(A\times B)\geqslant \mu(A\times \{x\}).$$
Since all the involved measures are probability measures, applying the previous inequality with $S\setminus A$ gives that 
$$
1-\nu\left(A\right)\geqslant \mu\left(S\times \{x\}\right)-\mu\left(A\times \{x\}\right),
$$
and since the marginal in the second coordinate is $\delta_x$, we got that 
$$
-\nu\left(A\right)\geqslant-\mu\left(A\times \{x\}\right)
$$
hence $\nu\left(A\right)\leqslant \mu\left(A\times \{x\}\right)\leqslant \mu\left(A\times B\right)$ which proves that 
$\mu\left(A\times B\right)=\nu\left(A\right)\delta_x\left(B\right)$. This identity is also true for $x\notin B$, and it can be extended to disjoint finite unions of cartesian products of elements of $\mathcal F$, which is sufficient to ensure the equality of two measures.
A: For any measurable set $B\subset S$, $\mu(S\times B)=\delta_x(B)=\mathbb{1}_B(x)$.  In particular, $\mu(S\times\{x\})=1$, and if $x\notin B$,
$$\mu(A\times B)=0=\nu(A)\delta_x(B),\qquad A\in\mathcal{F}$$
for $\mu(A\times B)\leq\mu(A\times(S\setminus\{x\})=0$.
Suppose now that $x\in B$. Then, $\delta_x(B)=1$, and $\mu(A\times(S\setminus B))=0$. Consequently,
$$\begin{align}
\nu(A)\delta_x(B)=\nu(A)&=\mu(A\times S)=\mu((A\times B) \cup (A\times(S\setminus B))\\
&=\mu(A\times B)+\mu(A\times(S\setminus B))=\mu(A\times B)
\end{align}$$
Putting things together, we have that
$$\mu(A\times B)=\nu(A)\times\delta_x(B),\qquad A,B\in\mathcal{F}$$
From this (why?) it follows that $\mu=\nu\otimes\delta_x$.
