Vtyjkyuk

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

Definitions: $L = L(X,mathbfX, mu)$ of integrable functions consists of all real-valued $mathbfX$-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 $f: X rightarrow mathbbR$ is $mathbfX$-measurable if $f^-1((alpha, infty)) $ are measurable sets for all real $alpha$.

If $f: X rightarrow mathbbR^+$ is $mathbfX$-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 $phi: X rightarrow mathbbR$ 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 $mathbfX$- 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?

This is not true, an easy counterexample is to let $A$ be a non-measurable subset of $[0,1]$ and define $$f : [0,1] rightarrow mathbbR, ; ; f(x) = begincases 1: & x in A; \ -1: & x notin A endcases.$$

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 $1$.

$vert f vert = f^+ + f^-$. 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 $int f^+dmu$ or $int f^-dmu$ were infinite, it wouldn't be the case that $int vert f vert dmu$ is finite.