Proof of Pre-image of Random Variable is $\mathcal{F}$-measurable

Question

I'd like to get some advice on proof of the following statements:

"Suppose $X$ is a random variable. Show that for every Borel subset $B \in \mathbb{R}$, the set $X^{-1}(B)$ is $\mathcal{F}$-measurable."

Then my proof is as follows:

For any Borel subset $B \in \mathbb{R}$, we can always specify a maximum value $b \in B$. Then by the definition of Random Variable, $\{\omega \mid X(\omega) \leq c\}$ is $\mathcal{F}$-measurable for every $c \in \mathbb{R}$. Since $b$ is also in $\mathbb{R}$, We can prove that $X^{-1}(B) \triangleq \{w|X(w) \in B\} = \{w|X(w) \leq b\}$ is also $\mathcal{F}$-measurable.

Is this proof valid?

Answer 1

It would be reasonable to include your definition of a RV. Many definitions actually define a RV to be measurable. From your question it appears that your definition of a random variable demands the measurability of only set of the Form $\{X \leq c\}$ for each $c$.

Your proof then is incorrect. You appear to want to be able to write the preimage of each measurable set as such a set, which is not possible. E.g. take $X:[0,1]\mapsto[0,1]$ with $X(x) = x$ and $B = [0,1/4]\cup[3/4,1]$.

Rather you should keep in mind that measurable sets form $\sigma$-Algebras. So if the preimage of each $(-\infty, c]$ is measureable, then any countable combination of (standard) set operations is measureable. As these permute with the preimage you immediately get that the preimage of each set in the $\sigma$-Algebra created by the sets of the Form $(-\infty, c]$ is $\mathcal F$-measurable.

And we know that one is in fact the Borel Algebra on $\mathbb R$, which proves the statement.

Answer 2

There is a problem. Let's consider the real line endowed with the Borel $\sigma$-algebra. Consider $B=(0,1]$ and the random variable $X: \mathbb{R} \to \mathbb{R}$, with $X(w)=w-\frac{1}{2}$ for $w\in{B}$ and $X(w)=0$ otherwise. We observe that $max\{B\}=1$ and we consider the set $\{w|X(w)\leq 1\}$. Then, $\{w|X(w)\leq 1\}=\mathbb{R}$, while $X^{-1}(B)=\{w|0< X(w)\leq 1\}=(\frac{1}{2},\frac{3}{2}]\subset \mathbb{R}$. Note that they are both Borel measurable sets, but they are not the same set. Thus the arguement does not hold.