The Doob-Dynkin lemma states that for two functions $X, Y \rightarrow \Omega$ the following two statements are equivalent:

  1. There is a Borel-measurable function $h:{R}^n\rightarrow R^n, f(X)=Y$.
  2. Y is $\sigma(X)$-measurable

Now, the german Wikipedia states:

The lemma shows why $\sigma$-algebras are considert the carriers of probabilitstic Information. If $Y$ is measurable by the $\sigma$-Algebra generated by X, then $Y$ cannot contain Information that is not already contained in $X$.

Question: Why does the Doob-Dynkin lemma show this?


1 Answer 1


Take a look at the English wikipedia:


There you will find that item (2) in your question is equivalent to $\sigma(Y)\subset\sigma(X)$. The heuristic explanation is more or less as follows. If $X:\Omega\rightarrow \mathbb{R}$ and $Y:\Omega\rightarrow \mathbb{R}$ are random variables, let $\omega\in\Omega$ be a point in the sample space (loosely, the result of an experiment). Suppose the scientist/analyst/statistician observes the values $X(\omega)$ and $Y(\omega)$. Through the value $X(\omega)$ she can then answer whether each of the events $E\in\sigma(X)$ has or has not occurred. For example, if $X(\omega)=2$ then from this value she can answer "yes!" in case someone asks her "is it true that $X(\omega)$ is a positive number?", which corresponds to the event $E=\left\{\omega': X(\omega')>0\right\}\equiv X^{-1}\left\{(0,+\infty)\right\}\in\sigma(X)$. Now if $\sigma(Y)\subset\sigma(X)$, or equivalently $Y=h\circ X$ for some measurable $h:\mathbb{R}\rightarrow\mathbb{R}$, then observation of the value $Y(\omega)$ brings no additional information to the statistician. Indeed, since every $G$ in $\sigma(Y)$ is also in $\sigma(X)$, the question "has the event $G\in\sigma(Y)$ occurred?" turns out to be equivalent to the question "has the event $G\in\sigma(X)$ occurred?". In particular $G$ will be of the form $X^{-1}(h^{-1}(B))$ for some Borel set $B\subset\mathbb{R}$

-- is it true that $Y(\omega)$ lies in $B$?

-- well, I know that $X(\omega)$ is not in $h^{-1}(B)$, so 'no'!


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