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All:

Say $f$ is a measurable (integrable, actually) function over the Lebesgue-measurable set $S$, with $m(S)>0$.

Now, since $m(S)>0$, there exists a non-measurable subset S' of $S$, and we can then write:

S=S'\cup (S\setminus S').

How would we then go about dealing with this (sorry, I don't know how to Tex an integral)

\int_S f\,d\mu=\int_{S'} f\,d\mu+ \int_{S\setminus S'}f\,d\mu? (given that S' and S\setminus S' are clearly disjoint)

Doesn't this imply that the integral over the non-measurable subset S' can be defined?

It also seems , using inner- and outer- measure, that if S' is non-measurable, i.e. $m^*, neither is S\setminus S'.

So I'm confused here. Thanks for any comments.

Edit: what confuses me here is this:

We start with a set equality $A=B$ (given as S=S'\cup (S-S'), so that $A=S$, B=S-S', from which we cannot conclude:

$\int_A f=\int_B f$ , it is as if we had $x=y+z$ , but we cannot then conclude, for any decomposition of $x$, that $f(x)=f(y+z)$.

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    Perhaps it's good to set $f=1$, you'll have measure, not integrals. If $A$ is nonmeasurable subset of measurable $x$, it is true that $\mu(X)=\mu(A \cup (X-A))$, but you cannot change this to $\mu(A)+\mu(X-A)$ since RHS is not defined.2011-12-16

2 Answers 2

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No, the Lebesgue integral over a nonmeasurable set is not defined. So you don't deal with that.

Yes, it is true that S\setminus S' is also nonmeasurable when $S$ is measurable and S'\subset S is nonmeasurable. Because if S\setminus S' were measurable, then S'=S\cap(S\setminus S')^c would be measurable.

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    @Rico: Arturo's analogies are very apt. AD's comments don't address your question. Of course if $A=B$, then $\int_A f=\int_B f$. In the last comment Arturo again points out the assumption you made that is not correct. $\int_{A\cup B}f=\int_{A\cup B}f$ is always true as long as $A\cup B$ is measurable (and this isn't saying much). But $\int_{A\cup B}f = \int_A f+\int_B f$ requires that $A$ and $B$ not only be disjoint, but they *must be measurable*. Another apt analogy had been posted yesterday but was deleted: $1=\lim_n 1=\lim_n [(n+1) -n]=\lim_n(n+1)-\lim_n n$ breaks down at the last "=".2011-12-16
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I am not entirely sure, and I don't have the time now to investigate it (nor do I have access to JSTOR), but I think something like your question may have been dealt with in the following paper:

R. L. Jeffery, Relative summability, Annals of Mathematics (2) 33 (1932), 443-459.

Also, try googling "Jeffery" along with the phrase "relative summability". Finally, the following paper might also be relevant, but I'm less sure: Othmar Zaubek, Über nicht meßbare Punktmengen und nicht meßbare Funktionen, Mathematische Zeitschrift 49 (1943-44), 197-218.

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    JSTOR link for those $w$ith access: http://www.jstor.org/pss/19685282011-12-17