# The Unapologetic Mathematician

## Measurable Functions on Pulled-Back Measurable Spaces

We start today with a possibly surprising result; pulling back a $\sigma$-ring puts significant restrictions on measurable functions. If $f:X\to Y$ is a function from a set into a measurable space $(Y,\mathcal{T})$, and if $g:X\to\mathbb{R}$ is measurable with respect to the $\sigma$-ring $f^{-1}(\mathcal{T})$ on $X$, then $g(x_1)=g(x_2)$ whenever $f(x_1)=f(x_2)$.

To see this fix a point $x_1\in X$, and let $F_1\subseteq Y$ be a measurable set containing $f(x_1)$. Its preimage $f^{-1}(F_1)\subseteq X$ is then a measurable set containing $x_1$. We can also define the level set $\{x\in X\vert g(x)=g(x_1)\}$, which is a measurable set since $g$ is a measurable function. Thus the intersection

$\displaystyle\{x\in X\vert g(x)=g(x_1)\}\cap f^{-1}(F_1)\subseteq X$

is measurable. That is, it’s in $f^{-1}(\mathcal{T})$, and so there exists some measurable $F\subseteq Y$ so that $f^{-1}(F)$ is this intersection. Clearly $f(x_1)\in F$, and so $f(x_2)$ is as well, by assumption. But then $x_2\in f^{-1}(F)\subseteq\{x\in X\vert g(x)=g(x_1)\}$, and we conclude that $g(x_2)=g(x_1)$.

From this result follows another interesting property. If $f:X\twoheadrightarrow Y$ is a mapping from a set $X$ onto a measurable space $(Y,\mathcal{T})$, and if $g:(X,f^{-1}(\mathcal{T})\to(Z,\mathcal{U})$ is a measurable function, then there is a unique measurable function $h:(Y,\mathcal{T})\to(Z,\mathcal{U})$ so that $g=h\circ f$. That is, any function that is measurable with respect to a measurable structure pulled back along a surjection factors uniquely through the surjection.

Indeed, since $f$ is surjective, for every $y\in Y$ we have some $x\in X$ so that $f(x)=y$. Then we define $h(y)=g(x)$, so that $g(x)=h(f(x))$, as desired. There is no ambiguity about the choice of which preimage $x$ of $y$ to use, since the above result shows that any other choice would lead to the same value of $g(x)$. What’s not immediately apparent is that $h$ is itself measurable. But given a set $M\in\mathcal{U}$ we can consider its preimage $\{y\in Y\vert h(y)\in M\}$, and the preimage of this set:

\displaystyle\begin{aligned}f^{-1}\left(\{y\in Y\vert h(y)\in M\}\right)&=\left\{x\in X\big\vert f(x)\in\{y\in Y\vert h(y)\in M\}\right\}\\&=\{x\in X\vert h(f(x))\in M\}\\&=\{x\in X\vert g(x)\in M\}\\&=g^{-1}(M)\end{aligned}

which is measurable since $g$ is a measurable function. But then this set must be the preimage of some measurable subset of $Y$, which shows that the preimage $h^{-1}(M)\subseteq Y$ is measurable.

It should be noted that this doesn’t quite work out for functions $f$ that are not surjective, because we cannot uniquely determine $h(y)$ if $y$ has no preimage under $f$.