# The Unapologetic Mathematician

## Composing Real-Valued Measurable Functions I

Now that we’ve tweaked our definition of a measurable real-valued function, we may have broken composability. We didn’t even say much about it when we defined the category of measurable spaces, because for most purposes it’s just like in topological spaces: given measurable functions $f:(X_1,\mathcal{S}_1)\to(X_2,\mathcal{S}_2)$ and $g:(X_2,\mathcal{S}_2)\to(X_3,\mathcal{S}_3)$ and a measurable set $M\in\mathcal{S}_3$, the measurability of $g$ tells us that $g^{-1}\in\mathcal{S}_2$, and the measurability of $f$ tells us that $(g\circ f)^{-1}(M)=g^{-1}(f^{-1}(M))\in\mathcal{S}_1$.

But now we’re treating $0$ a bit differently, and so we have to be careful. I say that if $\phi$ is a Borel measurable extended-real-valued function on the extended real line so that $\phi(0)=0$, and if $f$ is a measurable extended-real-valued function on a measurable space $(X,\mathcal{S})$, then the composition $\phi\circ f$ is measurable. Indeed, if $M$ is any Borel set, then we find

\displaystyle\begin{aligned}N(\phi\circ f)\cap(\phi\circ f)^{-1}(M)&=\{x\in X\vert\phi(f(x))\in M\setminus\{0\}\}\\&=\{x\in X\vert f(x)\in\phi^{-1}(M\setminus\{0\})\}\end{aligned}

Since $\phi(0)=0$, we can write

$\displaystyle\phi^{-1}(M\setminus\{0\})=\phi^{-1}(M\setminus\{0\}\setminus\{0\}$

And since $\phi$ is Borel measurable we know that $\phi^{-1}(M\setminus\{0\})$ is a Borel set. We can thus continue our calculation from above

\displaystyle\begin{aligned}\{x\in X\vert f(x)\in\phi^{-1}(M\setminus\{0\})\}&=\{x\in X\vert f(x)\in\phi^{-1}(M\setminus\{0\})\setminus\{0\}\}\\&=N(f)\cap\{x\in X\vert f(x)\in\phi^{-1}(M\setminus\{0\})\}\\&=N(f)\cap f^{-1}(\phi^{-1}(M\setminus\{0\}))\end{aligned}

which is measurable by the measurability of $f$

This is a sufficient, but far from a necessary condition. But it does allow us to bring in various useful functions in the place of $\phi$. For any positive real number $\alpha$ we have the function $x\mapsto\lvert x\rvert^\alpha$. If $\alpha$ is a positive integer, we have the function $x\mapsto x^\alpha$. These are all continuous, which implies that they’re Borel measurable, and they send $0$ back to itself. We conclude that any positive integral power of a measurable function is measurable, as is any positive power of the absolute value of $f$.

Of course, if $X$ itself is measurable as a subset of itself, then we need not tweak to our definition and we don’t need to add the requirement that $\phi(0)=0$. Also, the converse of this theorem is definitely not true; if $E$ is a non-measurable set, then the function $\chi_E-\chi_{E^c}$ is not measurable even though the absolute value $\lvert\chi_E-\chi_{E^c}\rvert=1$ is measurable.

It’s important to note here that we’re asking that $\phi$ be Borel measurable, because our definition of a measurable real-valued function is in terms of Borel sets in the target. Indeed, writing things out more thoroughly helps us see this: if $f:(X,\mathcal{S})\to(\mathbb{R},\mathcal{B})$ and $\phi:(\mathbb{R},\mathcal{L})\to(\mathbb{R},\mathcal{B})$ are measurable, then we can compose the functions on the underlying sets, but the target of $f$ isn’t the same measurable space as the source of $\phi$. There is thus no reason to believe that the composite would be measurable. And tomorrow I’ll give an example of just such a case.

May 4, 2010 - Posted by | Analysis, Measure Theory