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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I read some wikipedia pages, and have a question. I know how to construct a Vitali set (non-measurable), I understand relations and equivalence classes, but here is the problem.

The page says that $f = \chi_S - 1/2$ is not Lebesgue integrable but that $|f|$ is (constant). Here I assume that $$\chi_S := \left\{ \begin{array}{ll} 1, & x \in S\\ 0, & x \in S^c \end{array} \right.,$$ where $S$ is the nonmeasurable set of $\mathbb{R}$, the Vitali set.

I understand that $f = f^+ - f^-$ and $|f| = f^+ + f^-$, where $f^+ := \max(f,0)$ and $f^- := - \min(f,0)$.

So why is a nonmeasurable set suddenly measurable when the absolute value is taken?

I tried this:

$\int_{\mathbb{R}}\,|f|\,d\mu = \int_{\mathbb{R}}\,\bigg( \big(\chi_S-1/2\big)^+ + (\chi_S-1/2)^-\bigg)\,d\mu = \int_{\mathbb{R}}\,\big(\chi_S-1/2)^+\,d\mu + \int_{\mathbb{R}}\;\big(\chi_S - 1/2)^-\,d\mu$ $\leq \int_{\mathbb{R}}\;\chi_S{}^+\;d\mu+\int_{\mathbb{R}}\;(-1/2)^-\;d\mu+\int_{\mathbb{R}}\;\chi_S{}^-\;d\mu+\int_{\mathbb{R}}\;(-1/2)^-d\mu = \int_{\mathbb{R}}\;\chi_S{}^+d\mu+\int_{\mathbb{R}}\;\chi_S{}^-d\mu$,

a finite number apparently, while,

$\int_{\mathbb{R}}\,f\,d\mu = \int_{\mathbb{R}}\,\bigg( \big(\chi_S-1/2\big)^+ - (\chi_S-1/2)^-\bigg)\,d\mu = \int_{\mathbb{R}}\,\big(\chi_S-1/2)^+\,d\mu - \int_{\mathbb{R}}\,\big(\chi_S - 1/2)^-\,d\mu$ $\leq \cdots = \int_{\mathbb{R}}\;\chi_S{}^+d\mu-\int_{\mathbb{R}}\;\chi_S{}^-d\mu $

must be infinite? (The first equality holds only if both upper functions are Lebesgue integrable.) Or does this latter one fail just because it is undefined on $S$?

Thanks much!

P.S. This is a homework question suggesting to find a function $f$ that is not Lebesgue integrable, but whose $|f|$ is Lebesgue integrable. The question hinted at using $f = \chi_A - \chi_B$ on some subsets of $\mathbb{R}$. I tried tons of combinations (too many to type):

$\chi_A := 1$ if $x \in \mathbb{Q}$ and $-1$ if $x \in \mathbb{Q}^c$, with the opposite for $\chi_B$. I also tried intersecting $A$ and $B$ with $[0,1]$ so as to avoid an infinite measure that may have violated the Lebesgue integrability of $|f|$, too.

share|cite|improve this question
The function equals 1/2 on the non-measurable set and -1/2 off it. It isn't integrable, since it's not measurable; but the absolute value is constantly 1/2, thus integrable. – David Mitra Nov 13 '11 at 21:33
Just for finishing, it isn't measurable because the pre-image of $-1/2$ is not in a measurable set. – nate Nov 14 '11 at 23:21
up vote 7 down vote accepted

So why is a nonmeasurable set suddenly measurable when the absolute value is taken?

It is not measureable. The point is that the set $\{ x | |f(x)| = C\}$ is the union of two sets:

$$\{ x | |f(x)| = C\}= \{x | f(x)=C \} \cup \{x | f(x)=-C \} \,.$$

Now the union of two non-measurable sets can be measurable, and this is what lies behind this problem...

share|cite|improve this answer

I think that the formula $\int |f| d\mu = \int f^+ d\mu+\int f^- d\mu$ is valid only if both $f^+$ and $f^-$ are integrable. In the general case if $|f|$ is constant then it is integrable no matter if $f$ is. The trick is that you don't need to think of $|f|$ as an absolute value of some function, you can think about $|f|$ as a function itself.

share|cite|improve this answer
Good point. I'll change that just for the record. Thanks. – nate Nov 13 '11 at 22:43

The true issue is whether the set $S_c(f)$={x|f(x)>c}, for all reals c itself is measurable, not whether S itself is, tho of course, the two are related. Now, if f is not measurable, then there is some c for which $S_c(f)$ itself is not measurable (you can check for which specific c this happens, since f takes only finitely-many values). Now, |f| takes on different values than f, so that the sets $S_c(f)$ and $S_c(|f|)$ are different. Check to see that the problem set has dissapeared when considering |f|.

share|cite|improve this answer
Maybe to summarize (or, as one of my students once wrote, two sammurais), f is measurable if all $S_c$'s are. So, find a function f such that there is a negative c so that the only $S_c(f):={x:x>c}$ that is not measurable is a negative c. Then this obstacle to measurability of f will disappear with the absolute value composition. – C.O.Jones Nov 13 '11 at 22:04

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.