Exercise 1.7.20 (Product of Dirac measures and product of counting measures)

The trick is to use Exercise 1.7.9(i) and find a set $E\times F$ in the generating set $B_0=\{E\times F\mid E\in B_X,F\in B_Y\}$ such that $\delta (E\times F\setminus A)=0$. Since $E\times F$ is in the generating set, we can profit from the fact that the product of measures ${\delta }_x\cdot {\delta }_y$ and the product measure $\delta$ coincide on this domain. In other words, by additivity of $\delta$ we have

Hence,

From here on, we differentiate two mutually exclusive cases:

• either $(x,y)\in A$, in which case we must also have $(x,y)\in E\times F$ and so $\delta (E)={\delta }_x(E)\cdot {\delta }_y(E)=1$;
• or $(x,y)\notin A$, in which case we could, at first glance, have $(x,y)\in (E\times F)\setminus A$. But then, repeating the procedure from the previous part, but this time applied to $(E\times F)\setminus A$, we obtain a cover $E^{\prime }\times F^{\prime }\in B_0$ of $(E\times F)\setminus A$ such that $\delta (E^{\prime }\times F^{\prime })=\delta ([E\times F]\setminus A)=1$ as $(x,y)\in E^{\prime }\times F^{\prime }$ - a contradiction to $\delta ([E\times F]\setminus A)=0$ as per construction of $E\times F$.

Hence,

Now suppose, for the sake of contradiction, that $\#(A)\ne \left|A\right|$. We then have one of the following cases:

• $\left|A\right|>\#(A)$. In this case $\left|A\right|>|E\times F|$ although $A\subseteq E\times F$ - a contradiction.
• $\left|A\right|<\#(A)$. Then, we at least have some $(n,m)\in (E\times F)\setminus A$. Repeating the procedure, we can find a cover $E^{\prime }\times F^{\prime }\in B_0$ of $(E\times F)\setminus A$ with $\#\left(E^{\prime }\times F^{\prime }\right)=\#\left([E\times F]\setminus A\right)$. But $(n,m)\in [E\times F]\setminus A\subseteq E^{\prime }\times F^{\prime }$ - a contradiction to the fact that $|E^{\prime }\times F^{\prime }|=0$.