measurable function

Let \[(X,\Sigma)\] and \[(Y,T)\] be measurable spaces. A function \[f:X\to Y\] is said to be measurable if for every \[E\in T\], where \[E\] is also a subset of \[Y\], \[f^{-1}(E):=\left\{ x\in X\mid f(x)\in E \right\}\in \Sigma\].

In other words, for every measurable set \[E\] in sigma-algebra \[T\], define a collection \[f^{-1}(E)\] such that it contains every element \[x\] in set \[X\] that fulfils the condition \[f(x)\in E\]. If the set of \[x\] that fulfils this condition is also in \[\Sigma\], then \[f\] is said to be measurable.

We'll take probability theory as our model of reference as it is easier to understand. It's similar to measure theory. Just change the word "probability" to "measure" in the following passage and it will still make sense.

Define two measure spaces \[\left( X, \Sigma \right)\] and \[(Y,T)\], where \[\Sigma=\mathcal{P}(X)\] and \[T=\mathcal{P}(Y)\]. We can define \[\mu\] as a function that takes in some subset (a possible outcome) from the sigma-algebra \[\Sigma\] and spits out a probability between zero and one, i.e. \[\mu:\Sigma\to \left[ 0,1 \right]\]. Let's take a simple game with a fair dice as an example. The rules of the game dictate that if we roll an even, we get a dollar, if we roll an odd, we lose a dollar. A dice roll has six possible outcomes, so the set \[X=\left\{ 1,2,3,4,5,6 \right\}\]. Similarly, there are only two possible outcomes for our wallet, so the set \[Y=\left\{ -1,1 \right\}\].

It's trivial to see that we have a \[0.5\] chance of rolling an even or an odd. We can also say that we have a \[0.5\] chance of gaining and losing a dollar.

What we've done here is that we have transferred the possibility of an outcome from rolling a dice (\[X\]) to the possibility of gaining/losing a dollar (\[Y\]), i.e. \[f:X\to \left\{ -1,1 \right\}\implies f:X\to Y\]. This IS a measurable function. The possible outcomes (or events to be precise) to earning a dollar is \[f^{-1}(\left\{ 1 \right\})=\left\{ 2,4,6 \right\}\], and similarly the possible outcomes to losing a dollar is \[f^{-1}(\left\{ -1 \right\})=\left\{ 1,3,5 \right\}\]. We also know that the sets \[f^{-1}(\left\{ 1 \right\})\] and \[f^{-1}(\left\{ -1 \right\})\] are measurable, i.e. \[\mu \left( \left\{ 2,4,6 \right\} \right)=\mu \left( \left\{ 1,3,5 \right\} \right)=\frac{1}{2}\]. Thus, \[f\] fulfils the definition of a measurable function.