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I'm working on an interesting project regarding general measure theory. I'm wondering if you would be interested in hearing about it. Suppose $S$ is a set which is equinumerous to $\mathbb{R}$. We know that this means there exists a function $\phi$: $\mathbb{R}$ $\rightarrow$ $S$ that is one-to-one and onto. Now what I do is let $\mathcal{M}_\phi$ = {$E$ $\subset$ $S$ $\mid$ $\phi^{-1}$$(E)$ $\in$ $\mathcal{M}$}. I want to check that $\mathcal{M}_\phi$ is a $\sigma$-algebra. I'm clear with $\varnothing$ $\in$ $\mathcal{M}_\phi$. How do I show the corresponding properties for the complements of subsets of $\mathcal{M}_\phi$ and countable unions?

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What is $\mathcal M$ supposed to be? – echoone May 16 '11 at 23:42
@echoone: $\mathcal{M}$ is the $\sigma$-algebra of (Lebesgue) measurable sets. – Libertron May 16 '11 at 23:45
@Theo: There was no real problem except that I needed some clarification as to the notation used in order to show that this set was a $\sigma$-algebra. I see that it was the bijective property that makes things clearer. – Libertron May 17 '11 at 0:04
(I deleted my comment because it was basically saying the same thing as Calvin's answer below, but in a more brisk way) – t.b. May 17 '11 at 0:20
up vote 8 down vote accepted

Let $E \in \mathcal{M}$. Then $\overline{E} \in \mathcal{M}_\phi$ iff $\phi^{-1}(\overline{E}) \in \mathcal{M}$,and, since $\phi$ is a bijection, $\phi^{-1}(\overline{E}) = \overline{\phi^{-1}(E)}$ (as an element of $S$ is not in $E$ iff the corresponding $\phi^{-1}(s)$ is not in $\phi^{-1}(E)$). But, as $\mathcal{M}$ is a $\sigma$-algebra, $\overline{\phi^{-1}(E)} \in \mathcal{M}$. This proves the first part.

The second part follows with basically the same ease as the first- since $\phi$ is a bijection, it commutes with the relevant set-theoretic operations (complements and unions, in this case), so that the properties of $\mathcal{M}$ carry over.

Also, the result is expected, because the existence of such a bijection $\phi$ means that, from the point of view of elementary set theory (which is all that the definition of $\sigma$-algebra uses), $S$ and $\mathbb{R}$ are "the same" (i.e., isomorphic as sets), so it is to be expected that the set of correspondents of elements of $\mathcal{M}$ would be a $\sigma$-algebra too.

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