# $X$ is $\sigma(Y_{1}, ..., Y_{n})$-measurable iff $X = H(Y_{1}, ..., Y_{n})$ for some $H$ measurable

I am not sure if this is correct but I vaguely recall the professor mentioned something like this in class: Let $$(\Omega, \mathcal{F}, \mathbb{P})$$ be a probability space, and $$X, Y_{1}, ..., Y_{n}$$ random variables on it. Let $$\sigma(Y_{1}, ..., Y_{n})$$ be the sigma algebra generated by $$Y_{1}, ..., Y_{n}$$. Then $$X$$ is $$\sigma(Y_{1}, ..., Y_{n})$$-measurable iff $$X = H(Y_{1}, ..., Y_{n})$$ for some $$H$$ measurable.

Could someone explain the hard direction?

Here is an answer for the case where $$X$$ is a real random variable, and $$Y$$ a random variable taking values in any measurable space. In your situation, set $$Y = (Y_1,\ldots,Y_n)$$.

If $$X = 1_A$$ for some $$A \in \sigma(Y)$$ then the result follows from the definition of $$\sigma(Y)$$ : there exists some measurable subset $$B$$ such that $$A = [Y \in B]$$, so $$1_A=1_B(Y)$$.

By countable linear combinations, the result holds for every $$\sigma(Y)$$-measurable discrete real random variable.

Last, each $$\sigma(Y)$$-measurable real random variable $$X$$ can be written as a limit of $$\sigma(Y)$$-measurable discrete real random variables. For example, consider $$X_n = 2^{-n}\lfloor 2^X \rfloor$$. Then $$X_n = h_n(Y)$$ for some measurable function $$h_n$$. Then the function $$h = \limsup_{n \to +\infty} h_n$$ works.

• I have a question about $A$. For a single variable $Y_{1}$, we can say any element in $\sigma(Y_{1})$ has the form $Y_{1}^{-1}(B)$, where $B$ is a measurable set in the space where $Y_{1}$ takes values. But the definition of $\sigma(Y_{1}, ..., Y_{n})$ is the smallest sigma algebra that contains $\{Y_{i}^{-1}(B): i = 1, ..., n, B \text{ measurable}\}$. So a set $A$ from this sigma algebra may not have a form $Y_{1}^{-1}(B_{1}) \cap ... \cap Y_{n}^{-1}(B_{n})$ which in your case is $Y^{-1}(B)$. It may be a "generated" set so has a complicated explicit expression?
– Tom
Dec 14, 2022 at 15:21
• @Tom One can also introduce the random variable $Y := (Y_1,\ldots,Y_n)$ and check that $\sigma(Y) = \sigma(Y_1,\ldots,Y_n)$. Dec 14, 2022 at 15:31
• I get it. The details involve some product sigma algebra-related arguments about the space where $Y$ takes values. Thank you!
– Tom
Dec 14, 2022 at 16:05