When we define Lebesgue measure in $\mathbb{R}^n$, we usually make outer measure using rectangles and take induced measure by Caratheodory's method. Let $\mu_n$ be Lebesgue measure defined by this way.

Also, we can define Lebesgue measure by product measure. With $(\mathbb{R}^n,\mathcal{M}_n,\mu_n)$ and $(\mathbb{R}^m,\mathcal{M}_m,\mu_m)$, we define outer measure $(\mu_n\times\mu_m)^*$ with measureable Rectangles. Then also by Caratheodory method we get a measure on $\sigma(\mathcal{M}_n\times \mathcal{M}_m)$, and by completion we get Lebesgue measure exactly same as above.

Now, think about 'original' Caratheodory method. We can get measure on $(\mu_n\times\mu_m)^*$-measurable sets, not only on $\sigma(\mathcal{M}_n\times \mathcal{M}_m)$, if we give up uniqueness. My question is this: Is $(\mu_n\times\mu_m)^*$-measurable sets are exactly same as Lebesgue measurable set on $\mathbb{R}^{n+m}$? If yes, then measure defined on $(\mu_n\times\mu_m)^*$-measurable sets is exactly same as Lebesgue measure? If answer of second question is also yes, then is there exist measure defined on Lebesgue measurable set which is different from Lebesgue measure, coincide with Lebesgue measure at $\sigma(\mathcal{M}_n\times \mathcal{M}_m)$?


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