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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!

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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: $(2,3,7)=2(1,0,0)+3(0,1,0)+7(0,0,1)$.

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  • $\begingroup$ 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? $\endgroup$ – WhiteEyeTree Jan 12 at 18:06
  • $\begingroup$ @WhiteEyeTree Yes. $\endgroup$ – David C. Ullrich Jan 12 at 18:07

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