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In Folland's real analysis textbook, he says for two measure spaces $(X, \mathcal{M})$ and $(Y, \mathcal{N})$, and a function $f: X \rightarrow Y$ then $f$ is measurable if $f^{-1}(E) \in \mathcal{M}$ for all $E \in \mathcal{N}$.

He then goes on to discuss properties of integration on $L^+$, the set of all measurable functions $X \rightarrow [0,1]$.

Later, at the start of section 2.3, he starts extending these results to the real valued measurable functions. He says that $f$ (now any real valued function) is integrable if $\int |f| < \infty$. He then defines $L^1$ as the space of integrable real-valued functions.

My question is this, given these definitions, should I be taking all the functions in $L^1$ to be measurable? i.e. given an exercise which asks one to prove some function $f$ is measurable, is it sufficient to show $f$ is integrable? If so, why is measurability necessary in the definition in integrability?

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It is necessary, because without measurability of a function $f$, you can't define what the integral of $f$ should be.

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Integrable functions is a subclass of measurable functions $f$ (with $\int |f| < +\infty$), so by proving the former, you get the later.

To construct integrals, you first need to measure sets. These are the construction sites on which you build simple functions of (a) certain height(s). Then you use simple approximation for bounded functions. After that, we have the integral for positive functions and lastly for general measurable functions.

To measure a set is to "takes its length", but some sets can have (infinitely many) holes inside (e.g. Cantor set), so we construct measure with a few "idealistic" properties---but not all sets get along well with this function, so "measurability" is created to rule out these sets during the construction of integrals.

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