# Definition of lebesgue integral with respect to measure $\mu$ [duplicate]

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 $f$?

## marked as duplicate by Daniel FischerJun 24 '15 at 16:38

• 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. – Zach466920 Jun 24 '15 at 16:26
• 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 $$\int \chi_A\,d\lambda + \int \chi_{[0,1]\setminus A}\,d\lambda < 1.$$ – Daniel Fischer Jun 24 '15 at 16:29