# Show the existence of a Borel measurable function $\phi: \mathbb{R} \to \mathbb{R}$

Let $(X,S)$ be a measurable space and $f: X \to \mathbb{R}$ be $S-$measurable. Show that if $\psi:X \to \mathbb{R}$ is any $S_{0}-$measurable function, then there exists a Borel measurable function $\phi: \mathbb{R} \to \mathbb{R}$ such that $\psi=\phi \circ f$.

My try:

Case-1

If $\psi$ is a simple $S_0-$measurable function , then $\psi=\sum_{i=1}^{n}a_n\chi_{f^{-1}(E)}$ for some positive integer $n$, $a_i \in \mathbb{R}$ for each $i$ and $E_i \in B_{\mathbb{R}}$. Thus $\psi=(\sum_{i=1}^na_i\chi_{E_{i}})\circ f$.

Case-2

If $\psi$ is a non-negative $S_0-$measurable function, then there exists a increasing sequence of functions ${S_n}$ (simple measurable) such that $\psi(x)=\lim_{n \to \infty}S_{n}(x) \forall x \in X$. Since each $S_n(x)$ is a simple $S_0-$measurable function there exists a borel measurable function $\phi_n : \mathbb{R} \to \mathbb{R}$ such that $S_n=\phi_{n} \circ f$. This is where I am stuck. I am not sure if the limit of $\phi_n$ will exist or not and if it does the whether it is borel measurable or not. Moreover if the limit is $\phi(x)$, then can I conclude that $\psi(x)=\phi \circ f (x)$??

This will lead to the final conclusion as for any $\psi S_0-$measurable function $\psi^{+}$ and $\psi^{-}$ are non-negative $S_0$ measurable and I can conclude from the case-2.

Thanks for the help!!

• Recently came about the same problem. I had the very same progress and question haha. Anyway, are you sure about the expression of $ψ$ in case one? – Marko Karbevski Oct 19 '15 at 19:34

## 1 Answer

If $\psi$ is a non-negative $S_0$-measurable function, choose a monotone sequence $(\psi_n)$ of simple $S_0$-measurable functions with $\psi_n \to \psi$. By case 1, there are Borel functions $\phi_n$ with $\psi_n = \phi_n \circ f$ for each $n$. On the range of $f$, $(\phi_n)$ is monotone, as for $y \in f(X)$ there is some $x \in X$ with $y = f(x)$, giving $$\phi_n(y) = \phi_n f(x) = \psi_n(x) \le \psi_{n+1}(x) = \phi_{n+1}(y)$$ By changing $\phi_n$ outside $f(X)$ if necessary, e. g. replace $\phi_n$ with $\phi_n \cdot 1_{f[X]}$, we can assume that $(\phi_n)$ is monotone. Hence, there is some $\phi \colon X \to [0,\infty]$ with $\phi_n \to \phi$. As a limit of $S_0$-measurable functions, $\phi$ is $S_0$-measurable, moreover for $x \in X$: $$\psi(x) = \lim_n \psi_n(x) = \lim_n \phi_n\bigl(f(x)\bigr) = \phi\bigl( f(x)\bigr) = (\phi \circ f)(x)$$

• Don't you think we should compose rather than multiply? – tattwamasi amrutam Apr 2 '15 at 8:55
• By $\phi_n f$ I mean the composition, but I can add the $\circ$, if you want ... – martini Apr 2 '15 at 9:00
• Done :-) ${}{}$ – martini Apr 2 '15 at 9:00