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Questions about abstract measure and Lebesgue integral theory. Also concerns such properties as measurability of maps and sets.

9 votes
Accepted

Finiteness of Hausdorff measure of balls

As a counterexample, let $X$ be an infinite-dimensional normed space. For $\varepsilon<r/2$ it follows from Borsuk-Ulam that you need more than $n$ closed sets of diameter $\varepsilon$ to cover the i …
Martin Väth's user avatar
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7 votes
Accepted

Regarding a positive Lebesgue measure set in $\mathbb{R}^2$

The question was answered by Robert Israel 1995 on Usenet, essentially by the set mentioned in fedja's comment. The proof that this set has the required property is carried out in detail in Example 4. …
Martin Väth's user avatar
  • 1,869
7 votes
Accepted

To show a set is a set of positive Lebesgue measure in $ \mathbb{R}$

No. Every set $E$ without interior points (e.g. the complements of the rationals) has the property that $$\bigcap_{|t|<\varepsilon}(t+E)=\emptyset$$ for every $\varepsilon>0$. Indeed, for every $x\in …
Martin Väth's user avatar
  • 1,869
3 votes

Is there a standard way of defining the integral of an extended real function with respect t...

Once the integral over measurable bounded functions is defined, a “standard” way is this: For a nonnegative (measurable) function $f$ one takes the supremum of the integral over all bounded measurable …
Martin Väth's user avatar
  • 1,869
3 votes

On the intersection of two Orlicz spaces

It's many years ago that I read it, but I think that some of the most general interpolation type results for Orlicz spaces were contained in O’Neil, Richard, Integral transforms and tensor products on …
Martin Väth's user avatar
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3 votes

Why is Lebesgue measure theory asymmetric?

The asymmetry has only historical reasons. It is possible to develop Lebesgue theory (moreover, all extension theorems and thus the theory of product measures) from the “inner approach”. This was done …
Martin Väth's user avatar
  • 1,869
2 votes

Under what general conditions is the set $S := \left\{\int_{X}v(x)\pi(x)\,\mathrm{d}P(x) \mi...

I guess that if $v$ is $P$-integrable then the answer is positive, and actually the set is compact. Indeed, what you are looking for in this case is the compactness of the Aumann integral of the measu …
Martin Väth's user avatar
  • 1,869