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Let $\mu$ be a Borel measure and let $A$ be a $\mu$-measurable subset of $\mathbb{R}^\mathbb{N}$. I wish to approximate $A$ using a degenerate subset of $\mathbb{R}^\mathbb{N}$; more precisely, I wish to approximate $A$ by a set $B$ which is degenerate in the following sense: there exists an integer $n_0$ such that for each vector $(x_1,x_2,\ldots)$ in $B$ and for each vector $(y_1,y_2,\ldots)\in\mathbb{R}^\mathbb{N}$, the vector $(x_1,x_2,\ldots,x_{n_0},y_1,y_2,\ldots)$ is also in $B$.

By approximating I mean that given $\varepsilon>0$, I'll be able to find such $B$ for which $\mu(A\triangle B)<\varepsilon$.

For that to happen, I have to make sure that $A\triangle B$ is measurable as well; that will follow from requiring $B$ to be measurable. I have a feeling that if $B$'s projection on $\mathbb{R}^{n_0}$ will be identical to $A$'s projection (there), $B$'s measurability will follow, but I can't seem to prove that.

Anyway, this does not solve my entire problem, as I am still not sure why would I be able to approximate $A$ so (I am not even sure this is possible at all without further assumptions on $\mu$ or on $A$).

Bach
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2 Answers2

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The sets you call "degenerate" (they are known as measurable rectangles) form an algebra that generates the Borel-$\sigma$-algebra. So if $\mu$ is a finite Borel measure, the result follows from this answer of Davide or the way you did it.

But the result does not hold for infinite measures. Let $\lambda$ be the infinite product of Lebesgue measure on $M_1=[0,1]^\mathbb{N}$. Let $\nu$ be the measure on $M_2=\big(\mathbb{R}\backslash[0,1]\big)^\mathbb{N}$ that assigns infinite measure to every nonempty set. Define a Borel measure $\mu$ by $\mu(A)=\lambda(A\cap M_1)+\nu(A\cap M_2)$. Then every degenerate set has infinite measure, but $M_1$ has measure $1$.

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Michael Greinecker
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This may answer the question: call a set $C$ degenerate if there exists an integer $n_0$ such that for each vector $(x_1,x_2,\ldots)$ in $C$ and for each vector $(y_1,y_2,...)\in\mathbb{R}^\mathbb{N}$, the vector $(x_1,x_2,.\ldots,x_{n_0},y_1,y_2,\ldots)$ is also in $C$.

Denote by $\mathcal{B}$ the Borel sigma-algebra on $\mathbb{R}^\mathbb{N}$ and by $\mathcal{C}$ the set of all degenerate sets there. Let $\mathcal{B}_\mathcal{C}$ be the set of all sets in $\mathcal{B}$ which may be approximated by a set in $\mathcal{C}$. If we show that $\mathcal{B}_\mathcal{C}$ is a sigma-algebra, we're done.

Indeed, it is not empty, as $\mathbb{R}^\mathbb{N}$ is in it; if a set $B$ can be approximated by a set $C$, then $B^c$ can be approximated by $C^c$, which is also degenerate; so it is left to show that a countable union of sets from $\mathcal{B}_\mathcal{C}$ is there.

Let $B_1,B_2,\ldots\in\mathcal{B}_\mathcal{C}$, and let $B$ be their union. Let $C_1,C_2,\ldots\in\mathcal{C}$ be degenerate sets approximating $B_1,B_2,\ldots$ respectively, and let $C$ be their union. As the limit of the partial unions $\bigcup_{n=1}^k C_n$ is $C$ (as $k\to\infty$), we have that for large enough $k$, the partial union is a degenerate set approximating $C$, the latter approximating $B$.

Bach
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