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For example :

Consider the state space $\Omega = \mathbb{R}$, the $\sigma$-algebra, $\mathcal{F} = \{(-\infty, 0], (0, \infty), 0, \mathbb{R}\}$ and the random variable $X : \Omega \rightarrow \mathbb{R}$ defined by \begin{align*} X(\omega) = \begin{cases} 3 & \omega < 0\\ 5 & \omega \geq 0 \end{cases} \end{align*} Is $X$ $\mathcal{F}$-measurable?

Eugene Zhang
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lucas9876
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2 Answers2

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Measurability of a random variable $X$ is defined based on the inverse image.

Suppose $(\Omega, \mathcal{F}, P)$ is a probability space and $(\mathcal{X}, \mathcal{G})$ be a measurable space. Let $X$ be a random variable that is defined on this probability space by: $$X: (\Omega,\mathcal{F}) \rightarrow (\mathcal{X}, \mathcal{G})$$

Then, $X$ is said to be $\mathcal{F}-$measurable if for all $E \in \mathcal{G}, X^{-1}(E) \in \mathcal{F}$. Every pullback in the image of $X$ should be in the $\sigma-$algebra $\mathcal{F}$.

For your example, $\mathcal{G} = \big\{\{3\}, \{5\}, \{3,5\}, \emptyset\big\}$.

I think you mean to define $\mathcal{F} := \big\{(-\infty, 0), [0,\infty), \emptyset, \mathbb{R}\big\}\;$ based on your equality condition in your definition of $X$. In this way, you can verify that the above holds.

Eugene Zhang
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J. Stewart
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  • I also suspect that the definition of $\mathcal{F}$ was meant to include the empty set, instead of the number $0$, which is not a set. However, I suspect the intervals were meant to be $(-\infty, 0]$ and $(0, \infty)$, as written, which would make $X$ not $\mathcal{F}$-measurable. – Joe Mar 11 '21 at 16:01
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Just check if $\{\omega:X(\omega)<c\}\in \mathcal{F}$, $c \in \mathbb{R}$.

  1. $c<3$. In this case, since no $\omega$ satisfies, $\{\omega:X(\omega)<c\}=\varnothing\in \mathcal{F}$

  2. $3 \leqslant c<5$. In this case, $\{\omega:X(\omega)=3\}=(-\infty,0) \in \mathcal{F}$

  3. $c \geqslant 5$. In this case, $\{\omega:X(\omega)=5\}=[0,+\infty)=0\cup(0,+\infty)\in \mathcal{F}$

So $X$ is $\mathcal{F}$-measurable.

Eugene Zhang
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    Why is it not the case that for the 3rd condition we dont include the interval $(-\infty , 0)$ as its image is certainly less than $c$ which is greater than $3$? – Governor Oct 14 '21 at 18:17
  • Because in OP, $X(\omega) =3 :\text{for } \omega < 0 :\text{and } X(\omega) =5 :\text{for } \omega \geq 0$ – Eugene Zhang Oct 14 '21 at 18:41
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    I get that, but don't we want ${\omega : X(\omega) \le 5}$ for example, would that not include the sets for which $X(\omega) = 3$ and $X(\omega) = 5$ since both $5$ and $3$ are $\le 5$. – Governor Oct 14 '21 at 20:38
  • No, you misunderstood the $X(\omega)$. There must be a value for $X(\omega)$ for $\omega>0$ – Eugene Zhang Oct 14 '21 at 22:03