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One simple thing that can go wrong is purely related to the size of the space (polish spaces are all size $\leq 2^{\aleph_0}$). When spaces are large enough product measures become surprisingly badly behaved. Consider Nedoma's pathology: Let $X$ be a measure space with $|X| > 2^{\aleph_0}$. The diagonal in $X^2$ is not measurable.

We'll prove this by way of a theorem:

Let $U \subseteq X^2$ be measurable. $X$ U$can be written as a union of at most$2^{\aleph_0}$spaces of the form$A \times B$. Proof: We First note that we can find some countable collection$A_i$such that$U \subseteq \sigma(A_i \times A_j)$(proof: The set of$V$such that we can find such$A_i$is a sigma algebra containing the basis sets). For $x \in \{0, 1\}^\mathbb{N}$ define $B_x = \bigcap \{ A_i : x_i = 1 \} \cap \bigcap \{ A_i A_i^c : x_i = 0 \}$. Consider all sets which can be written as a (possibly uncountable) union of$B_x \times B_y$for some$y$. This is a sigma algebra and obviously contains all the$A_i \times A_j$, so contains$A$. But now we're done. There are at most$2^{\aleph_0}$of the$B_x$, and each is certainly measurable in$X$, so$U$can be written as a union of$2^{\aleph_0}$sets of the form$A \times B$. QED Corollary: The diagonal is not measurable. Evidently the diagonal cannot be written as a union of at most$2^{\aleph_0}$rectangles, as they would all have to be single points, and the diagonal has size$|X| > 2^{\aleph_0}$. 2 fixed latex One simple thing that can go wrong is purely related to the size of the space (polish spaces are all size$\leq 2^{\aleph_0}$). When spaces are large enough product measures become surprisingly badly behaved. Consider Nedoma's pathology: Let$X$be a measure space with$|X| > 2^{\aleph_0}$. The diagonal in$X^2$is not measurable. We'll prove this by way of a theorem: Let$U \subseteq X^2$be measurable.$X$can be written as a union of at most$2^\aleph_0$2^{\aleph_0}$ spaces of the form $A \times B$.

Proof: We can find some countable $A_i$ such that $U \subseteq \sigma(A_i \times A_j)$.

For $x \in {0, 1}^\mathbb{N}$ \{0, 1\}^\mathbb{N}$ define $B_x = \bigcap \{ A_i : x_i = 1 \} \cap \bigcap \{ A_i }$. \}$.

Consider all sets which can be written as a (possibly uncountable) union of $B_x \times B_y$ for some $y$. This is a sigma algebra and obviously contains all the $A_i \times A_j$, so contains $A$.

But now we're done. There are at most $2^\aleph_0$ 2^{\aleph_0}$of the$B_x$, and each is certainly measurable in$X$, so$U$can be written as a union of$2^{\aleph_0}$sets of the form$A \times B$. QED Corollary: The diagonal is not measurable. Evidently the diagonal cannot be written as a union of at most$2^\aleph_0$2^{\aleph_0}$ rectangles, as they would all have to be single points, and the diagonal has size $|X| > 2^{\aleph_0}$.

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One simple thing that can go wrong is purely related to the size of the space (polish spaces are all size $\leq 2^{\aleph_0}$). When spaces are large enough product measures become surprisingly badly behaved. Consider Nedoma's pathology: Let $X$ be a measure space with $|X| > 2^{\aleph_0}$. The diagonal in $X^2$ is not measurable.

We'll prove this by way of a theorem:

Let $U \subseteq X^2$ be measurable. $X$ can be written as a union of at most $2^\aleph_0$ spaces of the form $A \times B$.

Proof: We can find some countable $A_i$ such that $U \subseteq \sigma(A_i \times A_j)$.

For $x \in {0, 1}^\mathbb{N}$ define $B_x = \bigcap { A_i : x_i = 1 } \cap bigcap { A_i }$.

Consider all sets which can be written as a (possibly uncountable) union of $B_x \times B_y$ for some $y$. This is a sigma algebra and obviously contains all the $A_i \times A_j$, so contains $A$.

But now we're done. There are at most $2^\aleph_0$ of the $B_x$, and each is certainly measurable in $X$, so $U$ can be written as a union of $2^{\aleph_0}$ sets of the form $A \times B$.

QED

Corollary: The diagonal is not measurable.

Evidently the diagonal cannot be written as a union of at most $2^\aleph_0$ rectangles, as they would all have to be single points, and the diagonal has size $|X| > 2^{\aleph_0}$.