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 $x∈B$ ?

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.