Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 11 / 3 / ii / Solution

In the four functions theorem, take $\alpha =1_{\mathcal{A}}$, $\beta =1_{\mathcal{B}}$, $\gamma =1_{\mathcal{A}\vee \mathcal{B}}$ and $\delta =1_{\mathcal{A}\wedge \mathcal{B}}$, where the union and intersection of set families are the collections of all pairwise unions and intersections.

If $\alpha (A)\beta (B)=1$, then $A\cup B\in \mathcal{A}\vee \mathcal{B}$ and $A\cap B\in \mathcal{A}\wedge \mathcal{B}$, so the pointwise hypothesis holds. Otherwise its left side is zero. Summing these indicator functions gives