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Let $(X, \mathcal F)$ and $(Y, \mathcal G)$ be measurable spaces and $f:X\to Y$ be a measurable function. Let $\Gamma_f:X\to X\times Y$ be the map defined as $\Gamma_f(x)=(x, f(x))$.

Question. Is the image $\Gamma_f(X)$ a measurable subset of $X\times Y$, where the latter is equipped with the product $\sigma$-algebra?

The above question has answer in the afirmative if $X=Y=\mathbf R$ and the sigma algebras $\mathcal F$ and $\mathcal G$ are the Borel $\sigma$-algebras. This is because the map $X\times Y\to \mathbf R$ defined as $(x, y)\mapsto y-f(x)$ is a mesurable function whose zero-set is exactly $\Gamma_f(X)$.

But what about arbitrary measurable spaces?

caffeinemachine
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1 Answers1

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No. For a really simple example, let the $\sigma$-algebras $\mathcal{F}$ and $\mathcal{G}$ be trivial, consisting of just the whole space and the empty set. Then the product $\sigma$-algebra is also trivial, so no graph of a function is measurable if $X$ and $Y$ both have more than one point. However, every function $X\to Y$ is measurable.

For some less trivial examples, you can prove that if $A\subseteq X\times Y$ is measurable and we write $A_x=\{y\in Y:(x,y)\in A\}$, then $\{A_x:x\in X\}$ has cardinality at most $2^{\aleph_0}$. Indeed, the collection of sets $A$ with this property is easily verified to be a $\sigma$-algebra (using the fact that $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$), and it contains all rectangles, and so it contains the $\sigma$-algebra generated by measurable rectangles.

So, if the graph of a function is measurable, its image can contain at most $2^{\aleph_0}$ different points. In particular, for instance, if $X=Y$ is some measurable space of cardinality greater than $2^{\aleph_0}$, the graph of the identity map is not measurable.

Eric Wofsey
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