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 $x∈B$ ?


2 Answers 2


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.

  • $\begingroup$ 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 ? $\endgroup$
    – White
    Jan 16, 2019 at 14:08
  • 1
    $\begingroup$ @White I have edited in order to include more details. $\endgroup$ Jan 16, 2019 at 16:07
  • $\begingroup$ 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 $\mu.$ $\endgroup$
    – White
    Jan 16, 2019 at 16:35
  • $\begingroup$ @White Indeed, it seems that the opening poster knows that this is a property of product measure. $\endgroup$ Jan 16, 2019 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 $\mu=\nu\otimes\delta_x$.


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