Let M ⊂ R . Is M × M × Q ⊂ R 3 measurable? (Lebesgue measurable)I would say no, because by the definition the the Lebesgue measure, intuitively speaking, what it does is:We "approximate" our function by continuous functions with a compact support. Then we take the Riemann integral of that in each dimension.So if M is non-measurable it seems to me that then M × M × Q is too (so e.g. let M be the Vitali set).On the other hand, I think you could apply Tonelli Fubini: ∫ M ∗ ∫ M ∗ ∫ Q ∗ 1 M × M × Q = ∫ M ∗ ∫ M ∗ 0since the measure of Q is 0. So the above integral is also 0, so we can apply Tonelli, thus Fubini, and thus 1 M × M × Q is integrable and hence measurable.Is this correct? So my intuitive understanding is wrong, and instead it should be something like "since we can choose the order of the Riemann integrals, we can first do the one that results in zero, and then it doesn't matter that the other 'dimensions' aren't measurable anymore, since we have 0 everywhere"?

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Let   M  ⊂      R  . Is   M  ×  M  ×      Q    ⊂            R        3   measurable? (Lebesgue measurable)I would say no, because by the definition the the Lebesgue measure, intuitively speaking, what it does is:We &quot;approximate&quot; our function by continuous functions with a compact support. Then we take the Riemann integral of that in each dimension.So if M is non-measurable it seems to me that then   M  ×  M  ×      Q   is too (so e.g. let M be the Vitali set).On the other hand, I think you could apply Tonelli Fubini:      ∫    M    ∗        ∫    M    ∗        ∫                  Q              ∗        1          M      ×      M      ×              Q              =      ∫    M    ∗        ∫    M    ∗    0since the measure of       Q   is 0. So the above integral is also 0, so we can apply Tonelli, thus Fubini, and thus       1          M      ×      M      ×              Q             is integrable and hence measurable.Is this correct? So my intuitive understanding is wrong, and instead it should be something like &quot;since we can choose the order of the Riemann integrals, we can first do the one that results in zero, and then it doesn&#039;t matter that the other &#039;dimensions&#039; aren&#039;t measurable anymore, since we have 0 everywhere&quot;?

Dobermann82 · Accepted Answer

You&#039;re wrong.This set is Lebesgue measurable for any M.For example, if   M  ⊆  [  −  1  ,  1  ] is arbitrary, this is witnessed by the fact that   M  ×  M  ×      Q   is contained in       ⋂    m        U    m  , where       U    m    :=      ⋃    n        [    −  1  ,  1  ]  ×  [  −  1  ,  1  ]  ×  (      q    n    −  1      /    (  m      n    2    )  ,      q    n    +  1      /    (  m      n    2    )  ), where   (      q    n        )    n   is your favourite sequence of all rationals. Each       U    m   is an open set and its measure is bounded by             4              π        2                    3      m       (because       ∑          n      =      1        ∞        1          n      2        =            π      2        6  ), so its outer measure is zero. It easily follows that M is, in fact, measurable.If M is not contained in [−1,1], your set is covered by a countable union of similar sets.I&#039;m not sure where you get your intuition from, but it&#039;s wrong: Lebesgue measure does not have such a straightforward definition in terms of Riemann integrals.

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