dimension of vector space of measurable functions over finite set X

Suppose we have $$X=\{1,2,3\}$$ and $$M$$ a $$\sigma$$-algebra over X.

I've stumped across the question of finding the dimension of the vector space of all the $$M$$-measurable real functions over $$X$$ (e.g.$$\quad f:X \to \mathbb{R}$$). For the best of me i can't seem to wrap my head around this, even if i think it's rather intuitive.

My difficulty is coming up with a possible basis of functions for the vector space of functions from $$X$$ to $$\mathbb{R}$$. Once i get that checking for misurability is trivial in this case, but i don't know how to approach basis of functions.

Of course functions from $$X$$ to $$\mathbb{R}$$ have at most 3 distinct values. I thought of using as basis 3 separate functions, one for each value of $$X$$: $$e_i(i)=1\quad i=1,2,3$$.

But i don't think this cuts it since i can't come up with a way of writing every f via linear combination of these basis functions. Any hint on how to tackle this problem? Or maybe a general idea of how a base function should look like?

Thanks!

The dimension depends on $$M$$. If $$M=\{X,\emptyset\}$$ then measurable functions are constant, so the dimension is $$1$$. If, as you seem to be assuming, $$M$$ is the power set of $$X$$ then every function is measurable. In that case if you define $$e_x(y)=\begin{cases}1,&(y=x),\\0,&(y\ne x).\end{cases}$$then yes, the functions $$e_x$$ for $$x\in X$$ form a basis.
Hint for that: $$(2,3,7)=2(1,0,0)+3(0,1,0)+7(0,0,1)$$.
• If we assume $M=\{X, \emptyset\, \{1\}, \{2,3\}\}$ then this has dimension 2 with $(1,0,0), (0,1,1)$ as basis using your vectorial notation. Am i correct? – WhiteEyeTree Jan 12 at 18:06