There is a theorem in my textbook.
Theorem $5.1$
Let $f$ be a nonnegative function defined on a measurable set $E$. Then $\int_E f$ exists if and only if $f$ is measurable.
Notation
If $\int_E f$ exists and is finite, then $f$ is said to be integrable and denoted $f \in L(E)$.
I can attain the fact from theorem $5.1$ that, for a nonnegative $f$ definied on a measurable set $E$, $f$ is measurable if $f \in L(E)$.
Let there be no condition that $f$, defined on a measurable set $E$, is nonnegative. For this case, if $f \in L(E)$, then is $f$ measurable?
$f=f^+ - f^-$.
$f \in L(E)$ implies that $\int_E ({f^+ - f^-})$ is finite.
Since $\int_E f^+ - \int_E f^-$ is finite, both terms should be finite.
Thus, $f^+ \in L(E)$ and $f^- \in L(E)$.
By the theorem $5.1$, $f^+$ and $f^-$ are measurable.
The linear combination of the finite number of measurable functions is also measurable. Therefore, $f$ is measurable.
However, I cannot ensure that it is possible to proceed from the process $2$ to the process $3$.
I know the following theorems
If $\int_E f$ exists and $c$ is any real constant, then $\int_E (cf)$ exists and $\int_E (cf) = c\int_E f$.
if $f$ and $g$ are nonnegative and measurable, then $\int_E(f+g)=\int_E f+\int_E g$.
If $f, g\in L(E)$, then $f+g\in L(E)$ and $\int_E(f+g) = \int_E f + \int_E g$.
Although $f^+$ and $f^-$ are nonnegative, we cannot apply the second theorem since we don't know yet that they are measurable.
Also, since we don't know they are in $L(E)$, we cannot also apply the third theorem.
Thus, for any $f$ defined on a measurable set $E$, if $f\in L(E)$, then is $f$ measurable? If so, please let me know why~. Thank you.