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Question: Can every probability space $(X,\scr F,\mu)$ be $\sigma$-embedded in the completion of the space $[0,1]^K$ (equipped with a product of Lebesgue measure) for some set $K$?

Here, $f:\scr F\to \scr G$ is a $\sigma$-embedding of the measure space $(X,\scr F,\mu)$ into $(Y,\scr G,\nu)$ iff $f$ is one-to-one, and preserves measure, complements and countable unions.

Here's what I have so far:

Maharam's Theorem implies the measure (or probability) algebra of $X$ is $\sigma$-isomorphic to the measure (or probability) algebra of a disjoint union of countably many copies of $[0,1]^{K_n}$ (if $X$ has atoms, some of the $K_n$ will be empty), and of course the disjoint union $\sigma$-embeds in the measure algebra of $[0,1]^K$ for large enough $K$. Combining this with a lifting, we get map from $\scr F$ to the measurable sets of the completion of $[0,1]^K$ that preserves measure, complements and finite unions and is almost one-to-one (i.e., $f(A)=f(B)$ implies $\mu(A\Delta B)=0$).

So far I'm missing two ingredients: I have almost one-to-one instead of one-to-one and embedding instead of $\sigma$-embedding.

It looks like I can get the embedding to be one-to-one by making $K$ large enough that there be a null subset $N$ of $[0,1]^K$ of the same cardinality as $X$ and modifying the map $f$ from $\scr F$ to the measurable subsets of $[0,1]^K$ to $g(A)=(f(A)\backslash N)\cup h[A]$, where $h$ is a bijection between $X$ and $N$. This will be an embedding into the completion of $[0,1]^K$. But I want a $\sigma$-embedding...

I know next to nothing about measure algebras and the like, so I may be missing a completely standard and obvious thing.

Question: Can every probability space $(X,\scr F,\mu)$ be $\sigma$-embedded in the completion of the space $[0,1]^K$ (equipped with a product of Lebesgue measure) for some set $K$?

Here, $f:\scr F\to \scr G$ is a $\sigma$-embedding of the measure space $(X,\scr F,\mu)$ into $(Y,\scr G,\nu)$ iff $f$ is one-to-one, and preserves measure, complements and countable unions.

Here's what I have so far:

Maharam's Theorem implies the measure (or probability) algebra of $X$ is $\sigma$-isomorphic to the measure (or probability) algebra of a disjoint union of countably many copies of $[0,1]^{K_n}$ (if $X$ has atoms, some of the $K_n$ will be empty), and of course the disjoint union $\sigma$-embeds in the measure algebra of $[0,1]^K$ for large enough $K$. Combining this with a lifting, we get a map from $\scr F$ to ${\scr L}^K$, the measurable sets of the completion of $[0,1]^K$, that preserves measure, complements and finite unions and is almost one-to-one (i.e., $f(A)=f(B)$ implies $\mu(A\Delta B)=0$).

We can make the embedding be one-to-one by ensuring $K$ is large enough that there is a null subset $N$ of $[0,1]^K$ of the same cardinality as $X$ and replacing our almost one-to-one map $f:{\scr F}\to{\scr L}^K$ with $g(A)=(f(A)\backslash N)\cup h[A]$, where $h$ is a bijection between $X$ and $N$.

Then $g$ will be an embedding into the completion of $[0,1]^K$. But I want a $\sigma$-embedding.

Question: Is it true that every probability space $(X,\scr F,\mu)$ can be $\sigma$-embedded in the completion of the space $[0,1]^K$ (equipped with a product of Lebesgue measure) for some set $K$?

Here, $f:\scr F\to \scr G$ is a $\sigma$-embedding of the measure space $(X,\scr F,\mu)$ into $(Y,\scr G,\nu)$ iff $f$ is one-to-one, and preserves measure, complements and countable unions.

Here's what I have so far:

Maharam's Theorem implies the measure (or probability) algebra of $X$ is $\sigma$-isomorphic to the measure (or probability) algebra of a disjoint union of countably many copies of $[0,1]^{K_n}$ (if $X$ has atoms, some of the $K_n$ will be empty), and of course the disjoint union $\sigma$-embeds in the measure algebra of $[0,1]^K$ for large enough $K$. Combining this with a lifting, we get map from $\scr F$ in the measurable sets of $[0,1]^K$ that preserves measure, complements and finite unions and is almost one-to-one (i.e., $f(A)=f(B)$ implies $\mu(A\Delta B)=0$).

So far I'm missing two ingredients: I have almost one-to-one instead of one-to-one and embedding instead of $\sigma$-embedding.

It looks like I can get the embedding to be one-to-one by making $K$ large enough that there be a null subset $N$ of $[0,1]^K$ of the same cardinality as $X$ and modifying the map $f$ from $\scr F$ to the measurable subsets of $[0,1]^K$ to $g(A)=(f(A)\backslash N)\cup h[A]$, where $h$ is a bijection between $X$ and $N$. This will be an embedding into the completion of $[0,1]^K$. But I want a $\sigma$-embedding...

I know next to nothing about measure algebras and the like, so I may be missing a completely standard and obvious thing.

Question: Can every probability space $(X,\scr F,\mu)$ be $\sigma$-embedded in the completion of the space $[0,1]^K$ (equipped with a product of Lebesgue measure) for some set $K$?

Here, $f:\scr F\to \scr G$ is a $\sigma$-embedding of the measure space $(X,\scr F,\mu)$ into $(Y,\scr G,\nu)$ iff $f$ is one-to-one, and preserves measure, complements and countable unions.

Here's what I have so far:

Maharam's Theorem implies the measure (or probability) algebra of $X$ is $\sigma$-isomorphic to the measure (or probability) algebra of a disjoint union of countably many copies of $[0,1]^{K_n}$ (if $X$ has atoms, some of the $K_n$ will be empty), and of course the disjoint union $\sigma$-embeds in the measure algebra of $[0,1]^K$ for large enough $K$. Combining this with a lifting, we get map from $\scr F$ to the measurable sets of the completion of $[0,1]^K$ that preserves measure, complements and finite unions and is almost one-to-one (i.e., $f(A)=f(B)$ implies $\mu(A\Delta B)=0$).

So far I'm missing two ingredients: I have almost one-to-one instead of one-to-one and embedding instead of $\sigma$-embedding.

It looks like I can get the embedding to be one-to-one by making $K$ large enough that there be a null subset $N$ of $[0,1]^K$ of the same cardinality as $X$ and modifying the map $f$ from $\scr F$ to the measurable subsets of $[0,1]^K$ to $g(A)=(f(A)\backslash N)\cup h[A]$, where $h$ is a bijection between $X$ and $N$. This will be an embedding into the completion of $[0,1]^K$. But I want a $\sigma$-embedding...

I know next to nothing about measure algebras and the like, so I may be missing a completely standard and obvious thing.

Question: Is it true that every probability space $(X,\scr F,\mu)$ can be $\sigma$-embedded in the completion of the space $[0,1]^K$ (equipped with a product of Lebesgue measure) for some set $K$?

Here, $f:\scr F\to \scr G$ is a $\sigma$-embedding of the measure space $(X,\scr F,\mu)$ into $(Y,\scr G,\nu)$ iff $f$ is one-to-one, and preserves measure, complements and countable unions.

Here's what I have so far:

Maharam's Theorem implies the measure (or probability) algebra of $X$ is $\sigma$-isomorphic to the measure (or probability) algebra of a disjoint union of countably many copies of $[0,1]^{K_n}$, and of course the disjoint union $\sigma$-embeds in the measure algebra of $[0,1]^K$ for large enough $K$. Combining this with a lifting, we get map from $\scr F$ in the measurable sets of $[0,1]^K$ that preserves measure, complements and finite unions and is almost one-to-one (i.e., $f(A)=f(B)$ implies $\mu(A\Delta B)=0$).

So far I'm missing two ingredients: I have almost one-to-one instead of one-to-one and embedding instead of $\sigma$-embedding.

It looks like I can get the embedding to be one-to-one by making $K$ large enough that there be a null subset $N$ of $[0,1]^K$ of the same cardinality as $X$ and modifying the map $f$ from $\scr F$ to the measurable subsets of $[0,1]^K$ to $g(A)=(f(A)\backslash N)\cup h[A]$, where $h$ is a bijection between $X$ and $N$. This will be an embedding into the completion of $[0,1]^K$. But I want a $\sigma$-embedding...

I know next to nothing about measure algebras and the like, so I may be missing a completely standard and obvious thing.

Question: Is it true that every probability space $(X,\scr F,\mu)$ can be $\sigma$-embedded in the completion of the space $[0,1]^K$ (equipped with a product of Lebesgue measure) for some set $K$?

Here, $f:\scr F\to \scr G$ is a $\sigma$-embedding of the measure space $(X,\scr F,\mu)$ into $(Y,\scr G,\nu)$ iff $f$ is one-to-one, and preserves measure, complements and countable unions.

Here's what I have so far:

Maharam's Theorem implies the measure (or probability) algebra of $X$ is $\sigma$-isomorphic to the measure (or probability) algebra of a disjoint union of countably many copies of $[0,1]^{K_n}$ (if $X$ has atoms, some of the $K_n$ will be empty), and of course the disjoint union $\sigma$-embeds in the measure algebra of $[0,1]^K$ for large enough $K$. Combining this with a lifting, we get map from $\scr F$ in the measurable sets of $[0,1]^K$ that preserves measure, complements and finite unions and is almost one-to-one (i.e., $f(A)=f(B)$ implies $\mu(A\Delta B)=0$).

So far I'm missing two ingredients: I have almost one-to-one instead of one-to-one and embedding instead of $\sigma$-embedding.

It looks like I can get the embedding to be one-to-one by making $K$ large enough that there be a null subset $N$ of $[0,1]^K$ of the same cardinality as $X$ and modifying the map $f$ from $\scr F$ to the measurable subsets of $[0,1]^K$ to $g(A)=(f(A)\backslash N)\cup h[A]$, where $h$ is a bijection between $X$ and $N$. This will be an embedding into the completion of $[0,1]^K$. But I want a $\sigma$-embedding...

I know next to nothing about measure algebras and the like, so I may be missing a completely standard and obvious thing.