# 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.

• Could you explain the last conclusion in more detail. Im not sure how it works. You claim it follows if we apply the inequality with A = S\A ? Jan 16 '19 at 14:08
• @White I have edited in order to include more details. Jan 16 '19 at 16:07
• Thank you. I think you don't need that last sentence. The product measure is the unique measure satisfying $\nu \otimes \delta_x(A,B) = \nu(A) \delta_x(B)$ and this was proved for $\mu.$ Jan 16 '19 at 16:35
• @White Indeed, it seems that the opening poster knows that this is a property of product measure. Jan 16 '19 at 16:36

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 $$\mu=\nu\otimes\delta_x$$.