Number of measurable maps $f: \{1,2,3\} \rightarrow \{1,2,3\}$

Question

I'm asked to find the number of measurable maps $f:\Omega_1 \rightarrow \Omega_2$, where $\Omega_1 = \Omega_2 = \{1,2,3\}$.

Furthermore, we have that $\mathcal{F_1} = \mathcal{F_2} =\{\emptyset, \{1\}, \{2,3\}, \{1,2,3\}\}$ and that $(\Omega_1,\mathcal{F_1} ), (\Omega_2, \mathcal{F_2})$ are measurable spaces.

How can I proceed to find how many such maps exists? I'm not very good at measure theory since (by now) I've only seen a little and I don't know where I should start from. Any hints?

Answer 1

The number of measurable such maps is 15.

This is due to the fact that the preimage of any set in $\mathcal{F_2}$ must be in $\mathcal{F_1}$. Let's analyse which cases satisfy such condition:

1. We fix $f(1)=1$, then

• $f(2)=2$ & $f(3)=3$
• $f(2)=1$ & $f(3)=1$
• $f(2)=3$ & $f(3)=2$
• $f(2)=2$ & $f(3)=2$
• $f(2)=3$ & $f(3)=3$

These are the only maps that satisfy the above condition when $ f(1)=1$, hence they are measurable. Similarly:

2. $f(1)=2$, then

• $f(2)=2$ & $f(3)=3$
• $f(2)=1$ & $f(3)=1$
• $f(2)=3$ & $f(3)=2$
• $f(2)=2$ & $f(3)=2$
• $f(2)=3$ & $f(3)=3$

And

3. $f(1)=3$

• $f(2)=2$ & $f(3)=3$
• $f(2)=1$ & $f(3)=1$
• $f(2)=3$ & $f(3)=2$
• $f(2)=2$ & $f(3)=2$
• $f(2)=3$ & $f(3)=3$

The condition is always satisfied, therefore we have 3 situations having 5 measurable maps each, giving us a total of $$3\cdot5=15$$ measurable maps.