Can I say that every integrable function is measurable? Let $(X, \mathcal S, \mu)$ be a measure space. Let $\Bbb L$ be the collection of all $\mathcal S$-measurable functions and let $L_1(\mu)$ be the collection of all $\mu$-integrable functions i.e. the collection of all the functions $f : X \longrightarrow \Bbb R^*$ such that $\int f^+\ d\mu < +\infty$ and $\int f^-\ d\mu < +\infty,$ where $f^+$ and $f^-$ respectively denote the positive part and the negative part of the function $f.$ Now suppose that $f \in L_1(\mu).$ Can we say that $f \in \Bbb L?$ What I know is that if $f \in \Bbb L$ and $\int f^+\ d\mu, \int f^-\ d\mu < +\infty$ then $f \in L_1(\mu).$ Is the converse true?
Any help in this regard will be highly appreciated. Thanks in advance.
 A: Yes. Remember how the integral was defines. First for simple and of course measurable functions. measurability is necessary here or course. Then, for positive (measurable) functions, you basically construct a sequence of simple measurable functions using preimages of the positive functions. This is where -maybe silently- the measurability condition is used.
A: If $f$ is $\mu$-integrable, then it is not necessarily $\mathcal{S}$--measurable. It is however measurable with respect the completion of $\mathcal{S}$ under $\mu$.
Whether one uses Lebesgue-Charatheodory's cut condition (LC) to construct a measure or Daniell-Stones's functional approach (DS) to construct an  integral (and get a measure as a byproduct) there are a few important observations to make:

*

*In the process of building a measure (LC) or an integral (DS) starting from a simple object ($\sigma$-additive function on an algebra $\mathcal{A}$) or an elementary integral (linear functional on a Stone lattice $\mathcal{E}$) one obtains a larger class of objects which contained the simple object we started.


*The class of sets in the extension contains $\sigma(\mathcal{A})$ ( $\sigma(\mathcal{E})$ respectively).


*Any object $h$ in the extension, there is simple object $h'$ in $\sigma(A)$ ($\sigma(\mathcal{E})$ respectively) such that $h$ and $h'$ differ in a set of measure zero.
