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A measurable set is a pair $(X,\Sigma)$ where $X$ is a set and $\Sigma$ is a $\sigma$-algebra on $X$. The elements $U\in\Sigma$ will be considered as subsets $U\subseteq X$. A morphism of measurable sets from $(X,\Sigma)$ to $(X',\Sigma')$ is a function $f\colon X\to X'$ such that $f^{-1}(U')\in\Sigma$ for each $U'\in\Sigma'$. We can form a category $\mathcal{M}eas$ whose objects and morphisms are as above.

Question 1: What is a good reference for this category? In particular, I have sketched a proof that $\mathcal{M}eas$ has all finite limits, but I'd like to cite it in a book. I'd like also to peruse such a book.

For this MO post, I have a specific question about measures on measurable sets, namely whether the Beck-Chevalley condition holds. However, I'm not even sure the necessary structures exist to ask such a question.

Recall that a measure $\mu$ on $(X,\Sigma)$ is a function $\mu\colon\Sigma\to{\mathbb R}_+$ having certain properties, where ${\mathbb R}_+$ = $\{x\in{\mathbb R}\;|\;x\geq 0\}\cup\{\infty\}\ $ is the extended nonnegative real line. Given a morphism of measurable sets $f\colon (X,\Sigma)\to(X',\Sigma')$, one can push forward a measure $\mu$ on $(X,\Sigma)$ to give a measure $f_{\ast}\mu$ on $(X',\Sigma')$. There is a poset (and therefore category) structure on the set of measures, given by $\mu\leq\mu'$ iff $\mu(U)\leq\mu'(U)$ for each $U$, and measure push-forward $f_{\ast}$ is functorial with respect to that.

I've heard that there is no such thing as pull-back measures in general, but I wonder whether the push-forward $f_{\ast}$ has an adjoint, assuming we work with my made-up category of measures under $\leq$ as above. Given a measure $\mu'$ on $(X',\Sigma')$ and a subset $U\subseteq X$, I imagine defining $$(f^{\ast}\mu')(U)=\limsup_{U'\subseteq f(U)}\mu'(U').$$

Question 2: Will something like this work (i.e. be an adjoint to $f_{\ast}$)?

If something like this works, I'll assume (in order to ask the following question) that it works the way I think it will, i.e. that $f^{\ast}$ is a left adjoint to $f_{\ast}$.

Suppose that the diagram of measurable sets $(X,\Sigma_X)\xrightarrow{f}(Z,\Sigma_Z)\xleftarrow{g}(Y,\Sigma_Y)\ $ has as fiber product the measurable set $(W,\Sigma_W)$, together with morphisms $(X,\Sigma_X)\xleftarrow{h}(W,\Sigma_W)\xrightarrow{i}(Y,\Sigma_Y)\ $. I would like to draw the fiber product square here, but I don't know of a diagram package for mathjax. Given a measure $\mu$ on $(X,\Sigma_X)$ there are the following two ways to obtain a measure on $(Y,\Sigma_Y)$, and the following canonical morphism between them: $$g^{\ast}f_{\ast}\mu\to i_{\ast}h^{\ast}\mu.$$ The Beck-Chevalley condition holds iff this morphism is an isomorphism.

Question 3: Does the Beck-Chevalley condition hold, i.e. do we have $g^{\ast}f_{\ast}=i_{\ast}h^{\ast}$?

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Consider the case when X={a,b} consists of two points and X'={c} is a single point with μ' being the counting measure on X', i.e., μ'({c})=1. Your formula right before Question 2 gives fμ'({a})=fμ'({b})=fμ'({a,b})=1, thus fμ' is not additive, contradicting the definition of measure. –  Dmitri Pavlov May 1 '13 at 12:46

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