integration - bartle theorem 5.3 - absolute integrability

Question

Theorem 5.3: A measurable function $f$ belongs to $L$ if and only if $|f| $ belongs to $L$.

---

Definitions: $L = L(X,\mathbf{X}, \mu)$ of integrable functions consists of all real-valued $\mathbf{X}$-measurable functions $f$ defined on $X$, such that both the positive and negative parts $f^{+}, f^{-}$ of $f$ have finite integrals with respect to $\mu$.

A function $f: X \rightarrow \mathbb{R}$ is $\mathbf{X}$-measurable if $f^{-1}((\alpha, \infty)) $ are measurable sets for all real $\alpha$.

If $f: X \rightarrow \mathbb{R}^{+}$ is $\mathbf{X}$-measurable we define the integral with respect to $\mu$ to be the extended real number $$ \int f \, d \mu = \sup \int \phi \, d \mu ,$$ where supremum is taken over all simple functions $\phi: X \rightarrow \mathbb{R}$ such that $ 0 \le \phi \le f $. (The integral of simple function is then the usual one with $\mu$.)

---

What I don't see: $|f| \in L \Rightarrow f \in L$

We have $|f|^{+} = f^{+} + f^{-} , |f|^{-} = 0$ have finite integrals. But this does even not imply that $f^{+}$ and $f^{-}$ are $\mathbf{X}$- measurable. What am I missing?

---

EDIT: As given in the counter example by the comments, I believe the theorem should be reformulated as,

Let $f$ be a measurable function, then $f \in L \Leftrightarrow |f| \in L$

Is this correct?

Answer 1

This is not true, an easy counterexample is to let $A$ be a non-measurable subset of $[0,1]$ and define $$f : [0,1] \rightarrow \mathbb{R}, \; \; f(x) = \begin{cases} 1: & x \in A; \\ -1: & x \notin A \end{cases}.$$

Answer 2

The definition of integrable should include that $f$ is measurable. Otherwise, take $1-2\,\chi_E$ for any non-measurable set $E$, and its absolute value will be $1$.

Answer 3

$\vert f \vert = f^+ + f^-$. Then if $\vert f \vert$ belongs to $L$ it means that both $f^+$ and $f^-$ have finite integrals because we already know that the integral is linear (and the integral is defined over every measurable function and $f^+,f^-$ are measurable). Hence, if $\int f^+d\mu$ or $\int f^-d\mu$ were infinite, it wouldn't be the case that $\int \vert f \vert d\mu$ is finite.