Non measurable function verification

Question

Let $E$ be nonmeasurable subset of [0,1] and define $f:[0,1]$ as: $$f(x)=\begin{cases}-\frac1{2},\ \ x\in E\\0,\ \ \text{otherwise}\end{cases}$$

Then, I feel that both $f,|f|$ are clearly nonmeasurable, because preimage of the singleton sets $\{0\}\ \ ,\{\frac1{2}\}, \ \ \{\frac1{2}\}$ are non measurable. Am I right? I have two books which say that both $f,|f|$ are measurable in the answers section. I am confused. Any hints. Thanks beforehand.

Answer 1

We will prove the following useful and more general result, which provides an answer to your question:

Let $(X,\mathcal{A},\mu)$ be a measure space and $E\subseteq X$. Then: $$E\text{ is measurable }\Leftrightarrow \chi_E\text{ is measurable}$$ where, with $\chi_E$ we note the characteristic function of $E$.

(In all the following, $\mathbb{R}$ is considered to be equiped with the usual Lebesgue measure).

Proof:

($\Rightarrow$) Let $E$ be a measurable subset of $X$. Let also $a\in\mathbb{R}$ and let $A=\chi_E^{-1}((-\infty,a])$. Consider the following cases:

1. If $a\geq1$, then $A=X$, which is measurable.
2. If $0,\leq a<1$, then $A=X\setminus E$, which is measurable as the complementary set of $E$.
3. If $a<0$, then $A=\varnothing$, which is measurable.

So, in any case, $A$ is a measurable set, and, since $a$ was arbitrary, $\chi_E$ is measurable.

($\Leftarrow$) Let $\chi_E$ be measurable. Then, since $\{1\}$ is (Lebesgue) measurable, it is evident that $E=\chi_E^{-1}(\{1\})$ is measurable.

Notes:

1. It is clear that this proposition has a a straightforward implication that:

   $$E\text{ is not measurable }\Leftrightarrow\chi_E\text{ is not measurable}$$

2. It is also clear that the same applies for multiples of characteristic functions.

Taking into consideration all the above, your claim is right.