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I always see this word $\mathcal{F}$-measurable, but really don't understand the meaning. I am not able to visualize the meaning of it.

Need some guidance on this.

Don't really understand $\sigma(Y)$-measurable as well. What is the difference?

Kuba Helsztyński
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lakshmen
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2 Answers2

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Let $(\Omega,\mathcal{F},P)$ be a probability space, i.e. $\Omega$ is a non-empty set, $\mathcal{F}$ is a sigma-algebra of subsets of $\Omega$ and $P:\mathcal{F}\to [0,1]$ is a probability measure on $\mathcal{F}$. Now, suppose we have a function $X:\Omega\to\mathbb{R}$ and we want to "measure" the probability of $X$ belonging to some subset of $\mathbb{R}$. That is, we want to assign the probability to sets of the form $$\{X\in A\}:=X^{-1}(A)=\{\omega\in\Omega\mid X(\omega)\in A\}$$ for Borel sets $A\in\mathcal{B}(\mathbb{R})$. For this to make sense, we need to make sure that $\{X\in A\}\in\mathcal{F}$ for all $A\in\mathcal{B}(\mathbb{R})$, otherwise we can't assign a probability to it (recall that $P$ is only defined on $\mathcal{F}$).

Whenever $X:\Omega\to\mathbb{R}$ satisfies that $X^{-1}(A)\in\mathcal{F}$ for all $A\in\mathcal{B}(\mathbb{R})$ we say that $X$ is $(\mathcal{F},\mathcal{B}(\mathbb{R}))$-measurable or just $\mathcal{F}$-measurable when there is no chance of confusion. Thus, for a random variable $X$, it makes sense to assign the probability to any set of the form $\{X\in A\}$, and this defines the distribution of $X$: $$ P_X(A):=P(\{X\in A\}),\quad A\in\mathcal{B}(\mathbb{R}). $$ Note that a random variable is a synonym for an $\mathcal{F}$-measurable function.

If $Y:\Omega\to\mathbb{R}$ is a random variable, then $\sigma(Y)$ is, by definition, given as $$ \sigma(Y)=\sigma(\{Y^{-1}(A)\mid\ A\in\mathcal{B}(\mathbb{R})\}), $$ i.e. the smallest sigma-algebra containing all sets of the form $Y^{-1}(A)$. Another way of characterizing $\sigma(Y)$ is by saying that it is the smallest sigma-algebra we can put on $\Omega$ that makes $Y$ measurable.

Stefan Hansen
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    Thank you for this answer. As a suggestion / question, could you consider explaining the definition of the set ${X\in A}:=X^{-1}(A)$? Since $A$ is a Borel set, and $X^{-1}(A)$ is the inverse of the function (random variable) $X:\Omega \to \mathbb R$, it would seem like a "contradiction" to claim that $X^{-1}(A)$ could define a set with elements in $A$, i.e. ${X\in A}$, as opposed to a set in $\mathcal F.$ – Antoni Parellada Jan 30 '21 at 01:31
  • @Blank, there is no contradiction here. ${X\in A}$ is just notation for ${\omega\in\Omega\mid X(\omega)\in A}$ which is a subset of $\Omega$, not $A$. – Stefan Hansen Jan 30 '21 at 09:51
  • Thank you for clarifying this. It did have to be that way, since the answer is absolutely consistent. – Antoni Parellada Jan 30 '21 at 10:04
  • Is that an extra $\sigma$ at the end there? $\sigma(Y)={Y^{-1}(A)\mid\ A\in\mathcal{B}(\mathbb{R})}$? – Shuri2060 Sep 26 '22 at 09:59
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    @Shuri2060: Yes, as ${Y^{-1}(A)\mid A\in\mathcal{B}(\mathbb{R})}$ is not necessarily a sigma-algebra. – Stefan Hansen Sep 27 '22 at 06:00
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If $f\colon (X_1,\mathcal F_1)\to (X_2,\mathcal F_2)$, $f$ is $(\mathcal F_1,\mathcal F_2)$-measurable if for all $F_2\in\mathcal F_2$, $f^{-1}(F_2)\in\mathcal F_1$.

In some contexts we consider the case where $X_2$ is the real line and $\mathcal F_2$ the Borel $\sigma$-algebra. Then for short, we say that $f\colon X\to \mathbb R$ is $\mathcal F$-measurable if $f^{-1}(B)\in\mathcal F$ for each Borel subset $B$.

$\sigma(Y)$ is a $\sigma$-algebra, so the same definition applies.

Davide Giraudo
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