Let $X = \{a,b,c\}$. Find a measure space $(X × X × X, \mathcal{A}, \mu)$ so that $\int_{X × X × X} \,d\mu =1$.
This question has me somewhat stumped because I'm not sure how to approach integrating over a space made up of finite points instead of something like an interval or $\mathbb{R}$. I'm assuming something similar to the counting measure could work. Maybe a measure that is equal to 1 for all sets except the empty set? Would that still be equal to 1 when integrating over the space X × X × X?
If $(\Omega,\mathcal A,\mu)$ is a measure space then automatically we have $\Omega\in\mathcal A$ and $\int_{\Omega}d\mu=\mu(\Omega)$.
So in order to get $\int_{\Omega}d\mu=1$ it is enough that $\mu(\Omega)=1$.
To keep things as simple as possible we can choose for $\mathcal A=\{\varnothing,\Omega\}$. Then $\mathcal A$ is a $\sigma$-algebra and the function $\mu:\mathcal A\to[0,\infty]$ prescribed by $\varnothing\mapsto0$ and $\Omega\mapsto1$ is a measure on measurable space $(\Omega,\mathcal A)$. Then the triple $(\Omega,\mathcal A,\mu)$ can be recognized as a measure space with $\int_{\Omega}d\mu=1$.
The looks of set $\Omega\neq\varnothing$ are not really important. We can indeed go for $\Omega=X\times X\times X$ where $X=\{a,b,c\}$.