# If $Y$ is $\sigma(X)$-measurable, then there is a Borel function $f$ such that $f(X) = Y$

I'm trying to show that $$L_2(\Omega, \sigma(X), \mathbb R)$$ is equal to $$\Lambda$$ which is the set containing all random variables in $$L_2(\Omega, \mathcal F, \mathbb R)$$ that can be written as $$f(X)$$ for some Borel function $$f:\mathbb R \to \mathbb R$$. In doing so, I need to show, given $$Y \in L_2(\Omega, \sigma(X), \mathbb R)$$, there is a Borel function $$f:\mathbb R \to \mathbb R$$ such that $$Y = f(X)$$.

I'm stuck at proving the $$f$$ I construct is measurable. Could you elaborate on how to do so?

Let

• $$(\Omega, \mathcal F, \mathbb P)$$ be a probability space.

• $$L_2(\Omega, \mathcal F, \mathbb R)$$ the Lebesgue space of square-integrable random variables from $$(\Omega, \mathcal F, \mathbb P)$$ to $$(\mathbb R, \mathcal B(\mathbb R))$$.

• $$X \in L_2(\Omega, \mathcal F, \mathbb R)$$.

• $$\Lambda \subseteq L_2(\Omega, \mathcal F, \mathbb R)$$ containing all random variables in $$L_2(\Omega, \mathcal F, \mathbb R)$$ that can be written as $$f(X)$$ for some Borel function $$f:\mathbb R \to \mathbb R$$.

Then $$\Lambda = L_2(\Omega, \sigma(X), \mathbb R).$$

My attempt:

It's clear that $$\Lambda \subseteq L_2(\Omega, \sigma(X), \mathbb R)$$. Let $$Y \in L_2(\Omega, \sigma(X), \mathbb R)$$. Then we want to prove there is a Borel function $$f:\mathbb R \to \mathbb R$$ such that $$Y = f(X)$$.

For $$y_1, y_2 \in \mathbb R$$ such that $$y_1 \neq y_2$$, let $$A_1 =Y^{-1} (y_1)$$ and $$A_2 =Y^{-1} (y_2)$$. Then $$A_1 \cap A_2 = \emptyset$$. Because $$Y$$ is $$\sigma(X)$$-measurable, then there are Borel sets $$B_1, B_2$$ such that $$A_1 = X^{-1}(B_1)$$ and $$A_2 = X^{-1} (B_2)$$. It follows that $$X^{-1}(B_1) \cap X^{-1}(B_2) = X^{-1}(B_1 \cap B_2) =\emptyset$$. This means $$X(\omega) \notin B_1 \cap B_2$$ for all $$\omega \in \Omega$$. Thus $$X(A_1) \subseteq B_1 \setminus (B_1 \cap B_2)$$ and $$X(A_2) \subseteq B_2 \setminus (B_1 \cap B_2)$$. Hence $$X(A_1) \cap X(A_2) = \emptyset$$. As such, $$Y(\omega_1) \neq Y(\omega_2)$$ implies $$X(\omega_1) \neq X(\omega_2)$$.

Now we define the required $$f$$ by $$f(x) = Y(\omega)$$ if there is some $$\omega \in \Omega$$ such that $$x = X(\omega)$$ and $$f(x) = 0$$ otherwise. This construction is valid due to above justification. Let's prove that $$f$$ is Borel.

That result is due to Doob. Note that this is a property about measurable space $$(X,\mathcal{F})$$ only and no meausure is involved. Fix a random variable $$X$$ and let $$\mathcal{G}=\sigma(X)$$. Note that $$\mathcal{G}=\{X^{-1}(B)\mid B\in\mathcal{B}(\mathbb{R})\}$$. We go to show that for any $$\mathcal{G}$$-measurable random variable $$Y:\Omega\rightarrow\mathbb{R}$$, there exists a Borel function $$f:\mathbb{R}\rightarrow\mathbb{R}$$ such that $$Y=f\circ X$$.
Case 1: $$Y=1_{A}$$ for some $$A\in\mathcal{G}$$. In this case, there exists $$B\in\mathcal{B}(\mathbb{R})$$ such that $$A=X^{-1}(B)$$. For any $$\omega\in\Omega$$, $$1_{A}(\omega)=1$$ iff $$\omega\in A$$ iff $$X(\omega)\in B$$ iff $$1_{B}(X(\omega))=1$$. Hence, $$1_{A}=1_{B}\circ X$$. Note that $$1_{B}$$ is a Borel function.
Case 2: $$Y$$ is a $$\mathcal{G}$$-measurable simple functions. In this case, $$Y=\sum_{i=1}^{n}\alpha_{i}1_{A_{i}}$$ for some $$\alpha_{i}\in\mathbb{R}$$ and $$A_{i}\in\mathcal{G}$$. For each $$i$$, choose a Borel function $$f_{i}$$ such that $$1_{A_{i}}=f_{i}\circ X$$. Observe that $$Y=\sum_{i=1}^{n}\alpha_{i}f_{i}\circ X=f\circ X$$, where $$f=\sum_{i=1}^{n}\alpha_{i}f_{i}$$ is a $$\mathbb{R}$$-valued Borel function.
Case 3: $$Y:\Omega\rightarrow[0,\infty]$$ is a non-negative $$\mathcal{G}$$-measurable function. In this case, we can choose a sequence of $$\mathcal{G}$$-measurable random variables $$(Y_{n})$$ such that $$0\leq Y_{1}\leq Y_{2}\leq\ldots\leq Y$$ and $$Y_{n}(\omega)\rightarrow Y(\omega)$$ for each $$\omega\in\Omega$$. For each $$n$$, choose a Borel function $$f_{n}:\mathbb{R}\rightarrow\mathbb{R}$$ such that $$Y_{n}=f_{n}\circ X$$. Define $$f=\limsup_{n}$$. Note that $$f$$ is a $$[-\infty,\infty]$$-value Borel function (recall that limsup always exists). Let $$\omega\in\Omega$$. We have that $$Y(\omega)=\lim_{n}Y_{n}(\omega)=\lim_{n}f_{n}(X(\omega))=f(X(\omega))$$. Hence, $$Y=f\circ X$$. Note that if $$Y$$ is $$[0,\infty)$$-valued, $$f$$ can be choosen $$[0,\infty)$$-valued. For, define $$A=f^{-1}\left([0,\infty)\right)$$, which is a Borel set. Define $$g:\mathbb{R}\rightarrow[0,\infty)$$ by $$g(x)=f(x)1_{A}(x)$$. Clearly $$g$$ is also a Borel function. Let $$\omega\in\Omega$$, then from $$Y(\omega)=f(X(\omega))$$ and the fact that $$Y(\omega)\in[0,\infty)$$, we conclude that $$X(\omega)\in A$$. Therefore, $$g(X(\omega))=f(X(\omega))$$. Hence, $$Y(\omega)=g(X(\omega))$$, i.e., $$Y=g\circ X$$.
Case 4: $$Y:\Omega\rightarrow\mathbb{R}$$ is a $$\mathcal{G}$$-measurable function. Write $$Y=Y^{+}-Y^{-}$$, where $$Y^{+}=\max(Y,0)$$ and $$Y^{-}=\max(-Y,0)$$. Choose Borel functions $$f_{1},f_{2}:\mathbb{R}\rightarrow[0,\infty)$$ such that $$Y^{+}=f_{1}\circ X$$ and $$Y^{-}=f_{2}\circ X$$. Define $$f=f_{1}-f_{2}$$, which is a Borel function. Let $$\omega\in\Omega$$, then we have $$Y(\omega)=Y^{+}(\omega)-Y^{-}(\omega)=f_{1}(X(\omega))-f_{2}(X(\omega))=f(X(\omega))$$. Hence, $$Y=f\circ X$$.