# Understanding $a_i$'s in simple functions in measure theory

In search for help here. Will appreciate any support from you all. Thanks in advance.

So, I have already checked a couple of questions I found here:

Understanding Simple Functions

and:

Representations of Simple Functions

I think I understand now what a simple function represents.

However, I need some help understanding the meaning of the $a_i$'s.

Lets say, we have a measurable space $(\Omega,\mathcal{B})$, then a mapping X, (which is basically a random variable X), corresponds to:

$X(\omega): (\Omega,\mathcal{B})\mapsto (\mathbb{R},\mathcal{B}(\mathbb{R}))$.

Then, we say that the range of $X(\omega)$ is finite, correct?, then for given finite partition of $\Omega$, such that we have $n$ subsets $A_i \in \Omega$:

$\Omega = \bigcup_i^n A_i$ for $A_i \cap A_j = \emptyset$, and $i\neq j$.

Thus, we can represent $X(\omega)$ as:

$X(\omega) = \sum_i a_i \cdot \mathbb{1}_{A_i}$

What I think:

Well, my understanding is that the mapping $X(\omega) = a_i$

Since $a_i \in \mathcal{B}(\mathbb{R})$, the meaning of $a_i$ is the value of the set in the new measurable space $(\mathbb{R}, \mathcal{B}(\mathbb{R}) )$.

Then, we can also have: $A_i = \{ \omega \in \Omega: X(\omega) = a_i\} = X^{−1}({a_i})$

Then $a_i$ is the actual set in the new space. Actually, a real value.

Please, help me understanding this better!! I'm new on the field and lack a lot of background and proper math vocabulary use.

Why is it that the $a_i$'s seem to be like a weight for each indicator function??

$$X(\omega) = \sum_i a_i \cdot \mathbb{1}_{A_i}$$
This just says that X takes the value $a_i$ on the set $A_i$. If $\omega \in A_i$ for some i, then$X_{\omega}=a_i$ because for all $j \neq i$, $\mathbb{1}_{A_j} = 0$.
Yes, it kinda looks like a weighting function. If $\sum_i a_i=1$, then the expression is indeed the weighted sum (as we normally understand that term) those characteristic functions.
Note also that if $\sum_i \mu(A_i)=1$, then the integral of X is the weighted sum, (by weights $\mu(A_i)$, of the $a_i$s.
• Thanks for your reply. What is the meaning of the $\mu_i$ you just wrote? – kentropy Nov 15 '17 at 18:09
• I meant the measure of the sets $A_i$. Please see edits. – Mathemagical Nov 15 '17 at 22:30