I read some wikipedia pages, http://en.wikipedia.org/wiki/Absolute_convergence 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.
Best Answer
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...