The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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August 3, 2010 - Posted by | Analysis, Measure Theory

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