Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.


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 $f(x)=f(y+z)$.

share|improve this question
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. –  Arturo Magidin Dec 16 '11 at 6:16
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 $S'$. –  Arturo Magidin Dec 16 '11 at 6:18
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 –  AD. Dec 16 '11 at 6:43
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. –  sdcvvc Dec 16 '11 at 22:18

2 Answers 2

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.

share|improve this answer
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. –  Rico Dec 16 '11 at 6:32
@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. –  Arturo Magidin Dec 16 '11 at 6:37
@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. –  Jonas Meyer Dec 16 '11 at 6:38
@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 $f(y+z)=f(y)+f(z)$". –  Arturo Magidin Dec 16 '11 at 16:41
@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 "=". –  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.

share|improve this answer
@Jonas Meyer: Thanks. I've made the correction. –  Dave L. Renfro Dec 16 '11 at 20:48
JSTOR link for those with access: jstor.org/pss/1968528 –  Jonas Meyer Dec 17 '11 at 7:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.