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In Rudin's Real and Complex Analysis, the Lebesgue integral is defined as:

L et $(X,m,\mu)$ be a measure space, where $X$ is a set, $m$ is a $\sigma$ algebra on $X$ and $\mu$ is a measure. Then, if $f:X \to [0,\infty]$ and $E \in m$, we define $$\int_E f d\mu = \sup \int_E s d\mu \tag{1}$$ where the supremum is taken over all simple functions $s, 0 \leq s \leq f$

I do not have much background in measure theory, and I am wondering why we assume $f$ to be measurable to define its integral.

EVEN IF $f$ is not a measuarable function, the above definition (1) would still be well-defined.

Why do we define integral only for measurable $f$?

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marked as duplicate by Daniel Fischer Jun 24 '15 at 16:38

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  • $\begingroup$ Heuristically, trying to integrate a function with respect to measure, with a function that isn't measurable, is like trying to solve $x+1.2=7$ with the constraint that x is rational, given the fact that x is integer. $\endgroup$ – Zach466920 Jun 24 '15 at 16:26
  • $\begingroup$ I think this has been asked before. While $(1)$ is well-defined for any non-negative $f$, the integral only has nice properties when it is restricted to measurable functions. If, for example, you take a non-measurable subset $A\subset [0,1]$, then $\chi_A + \chi_{[0,1]\setminus A} = \chi_{[0,1]}$, but $$\int \chi_A\,d\lambda + \int \chi_{[0,1]\setminus A}\,d\lambda < 1.$$ $\endgroup$ – Daniel Fischer Jun 24 '15 at 16:29
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What you have there is the lower integral. There's also a similar upper integral. If they are equal, then the function is said to be integrable. So yes, your supremum is always defined, and that's a very good thing indeed. But it doesn't have nice sigma-additivity properties unless the upper and lower integrals are equal. That's why you need the integrability. There is a similar consideration for measures of sets, where there are inner measures and outer measures, and you get lots of nice properties if they are equal. Something similar also occurs with Riemann, Darboux and Stieltjes integrals.

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