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. $m^*<m_*$, 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 $f(x)=f(y+z)$.

  • $\begingroup$ If $S$ is measurable, and $S'$ is not measurable, then $S-S'$ is not measurable. So both $\int_{S'}fd\mu$ and $\int_{S-S'}fd\mu$ are integrals over nonmeasurable sets. You can't define $\int_{S'}fd\mu$ as $\int_Sfd\mu - \int_{S-S'}fd\mu$, because the last integral is not defined either. $\endgroup$ – Arturo Magidin Dec 16 '11 at 6:16
  • $\begingroup$ The fact that if $S$ is measurable and $S'$ is not measurable then $S-S'$ is not measurable follows from the fact that the $\sigma$-algebra of measurable sets is closed under under differences: $S'= S-(S-S')$, so if $S$ and $S-S'$ are measurable, then so is $S'$. $\endgroup$ – Arturo Magidin Dec 16 '11 at 6:18
  • $\begingroup$ I would recommend you to take a look at Banach–Tarski paradox, that give a glimpse of how bad non-measurable sets might be: en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox $\endgroup$ – AD. Dec 16 '11 at 6:43
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    $\begingroup$ 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. $\endgroup$ – sdcvvc Dec 16 '11 at 22:18

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.

  • $\begingroup$ I understand, but then we have the curious situation:S=S'U(S\S'), i.e., both sets are equal, but the integral over S exists and the integral over its equal counterpart does not. $\endgroup$ – Rico Dec 16 '11 at 6:32
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    $\begingroup$ @Rico: $3+1$ is evenly divisible by $2$, but neither $3$ nor $1$ are. Neither of the series $\sum\frac{1}{n}$ nor $\sum\frac{1}{n+1}$ converge, but $\sum(\frac{1}{n}-\frac{1}{n+1})$ converges to $1$... You can add two functions that are discontinuous everywhere and get a function that is continuous everywhere... Etc, etc. etc. etc. It's only "curious" if you think that any way of breaking up a good thing must result in two good things. $\endgroup$ – Arturo Magidin Dec 16 '11 at 6:37
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    $\begingroup$ @Rico: No, if $A=B$ and $A$ is measurable, the $\int_A f$ and $\int_B f$ both exist and are equal (this is not saying much). So if $S$ is measurable and $S'\subset S$, then $\int_S f=\int_{S'\cup(S\setminus S')}f$ is true, no problem. But $\int_{S'} f$ does not exist, so you cannot break up the integral in the way you indicated. If $A$ and $B$ are measurable and disjoint, then $\int_{A\cup B}f=\int_A f+\int_B f$. If $A$ is nonmeasurable, then at least the right-hand side of the previous equation is undefined. $\endgroup$ – Jonas Meyer Dec 16 '11 at 6:38
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    $\begingroup$ @Rico: You are incorrect in your analogy, because you are not trying to conclude that $\int_Sf = \int_{S'\cup (S-S')}f$. That equality is true. What fails is the claim that $\int_{S'\cup(S-S')}f = \int_S'f + \int_{S-S'}f$. There you are not trying to equation two integrals over identical sets, you are trying to decompose the integral. In your functional example, you can go from $x=y+z$ to $f(x)=f(y+z)$. What you cannot do is then say "and $f(y+z)=f(y)+f(z)$". $\endgroup$ – Arturo Magidin Dec 16 '11 at 16:41
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    $\begingroup$ @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 "=". $\endgroup$ – Jonas Meyer Dec 16 '11 at 16:47

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.

  • $\begingroup$ @Jonas Meyer: Thanks. I've made the correction. $\endgroup$ – Dave L. Renfro Dec 16 '11 at 20:48
  • $\begingroup$ JSTOR link for those with access: jstor.org/pss/1968528 $\endgroup$ – Jonas Meyer Dec 17 '11 at 7:02

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