Can a countable union of non-measurable sets of reals be measurable? For instance, can we partition $\mathbb{C}$ into countably many disjoint non-measurable sets?
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Sure. Just take any non-measurable set and its complement. For example, let $V$ be a Vitali set, which is known to be non-measurable. We have $$ \mathbb R = V \cup V^\complement. $$
Ayman Hourieh
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Ah yes, silly me! – user48900 Nov 03 '13 at 10:39
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Isn't the mesure of $V^C = \infty$? Considering that Vitali set is constructed in an interval. Do you first fill the space with copies of Vitali set? – Karolis Juodelė Nov 03 '13 at 10:49
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@KarolisJuodelė No. If $V^\complement$ were measurable, its complement $V$ would be measurable too. – Ayman Hourieh Nov 03 '13 at 10:51
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@AymanHourieh, clearly, ${x>2} \subset V^\complement$ and $m ({x>2}) = \infty$. Is there some contradiction in saying that $\infty = m ({x>2}) \leq m(V^\complement) = \infty$? – Karolis Juodelė Nov 03 '13 at 10:59
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@KarolisJuodelė When do you say a set is Lebesgue measurable? This needs to be based on a $\sigma$-algebra. If you say $m(V^\complement) = \infty$, this implies that $V^\complement$ and hence $V$ are measurable. Contradiction. – Ayman Hourieh Nov 03 '13 at 11:03
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1@KarolisJuodelė Have a look at this question. We can use your reasoning to assign an inner measure to $V^\complement$, but not a measure! – Ayman Hourieh Nov 03 '13 at 11:13
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@AymanHourieh, $m_{inner}(V^\complement) = m_{outer}(V^\complement) = \infty$. Doesn't this imply $m(V^\complement) = \infty$? – Karolis Juodelė Nov 03 '13 at 11:49
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@KarolisJuodelė This construction only works for bounded sets. For unbounded sets, one considers the intersection with bounded intervals. This is where your equality fails for $V^\complement$. Have a look here. – Ayman Hourieh Nov 03 '13 at 12:31
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@AymanHourieh, I see. Thanks. – Karolis Juodelė Nov 03 '13 at 12:55
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I can clearly say that in generally, countable union of non measurable sets is not non measurable. We know that union of every Vitali sets defined on [0,1] closed intervals is [0,1] and rational numbers are countablr so that we can choose some vitali sets countably which union is [0,1]
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