(why) does an integrable function have to be measurable?

Question

In this definition, it says that for a function $u$ to be $\mu$-integrable, it also has to be measurable:

However, Here's what I am wondering:

Take the Lebesgue integral of $u^+$ (the same will apply to $u^-$). This Lebesgue integral is the limit of simple functions $f_i$ on $X$. Assume that we have a sequence of simple functions on $X$ that are all measurable and as $n\to \infty, f_i\to u^+$. Can't we then say: As we let $n\to \infty$, the integral of $f_n$ will go to the integral of $u^+$ even if $u^+$ is not measurable. We don't need $u^+$ to be measurable, so long as all the $f_i$'s are measurable.

Is this correct?

Answer 1

The limit of a sequence of measurable functions must be measurable.

For the Vitali non-measurable set $C$, then consider the non-measurable function $u=\chi_{C}$, in this case, $u^{+}=\chi_{C}$, so there is no any sequence $(f_{n})$ of measurable functions such that $f_{n}\rightarrow\chi_{C}$ a.e.

Answer 2

Actually you are missing one thing, a sequence of measurable function converges to a measurable function Hence so does the sequence of measurable simple functions therefore we require the functions to be measurable in the definition of Integrability. Hope it works