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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. $U$ can be written as a union of at most $2^{\aleph_0}$ spaces of the form $A \times B$.

Proof: 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 $B_x = \bigcap \{ A_i : x_i = 1 \} \cap \bigcap \{ A_i^c : x_i = 0 \}$.

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 $|X| > 2^{\aleph_0}$.

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 measurable space with $|X| > 2^{\aleph_0}$. Then the diagonal in $X^2$ is not measurable.

We'll prove this by way of a theorem:

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

Proof: First note that we can find some countable collection $(A_i)_{i\ge 0}$ of subsets of $X$, such that $U \subseteq \sigma(\{A_i \times A_j:i,j\ge 0\})$, where $\sigma(\cdot)$ denotes the $\sigma$-algebra generated by the given subsets (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 $B_x = \bigcap \{ A_i : x_i = 1 \} \cap \bigcap \{ A_i^c : x_i = 0 \}$.

Consider all subsets of $X^2$ 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 $U$.

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 $2^{\aleph_0}$ subsets 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 $|X| > 2^{\aleph_0}$.

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 $|X| > 2^{\aleph_0}$.

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. $U$ can be written as a union of at most $2^{\aleph_0}$ spaces of the form $A \times B$.

Proof: 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 $B_x = \bigcap \{ A_i : x_i = 1 \} \cap \bigcap \{ A_i^c : x_i = 0 \}$.

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 $|X| > 2^{\aleph_0}$.

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 $|X| > 2^{\aleph_0}$.

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 $|X| > 2^{\aleph_0}$.