Why define the Lebesgue-Integral just for measurable functions?

Question

Usually, the Lebesgue integral, for example on Wikipedia, is defined for non-negative measureable functions as $$ \int_E f \, d\mu := \sup\left\{ \int_E s \, d\mu : 0 \le s \le f, s \text{ simple } \right\}. $$ But why suppose that $f$ should be measureable? This is not used in the definition here, so we could define this supremum for arbitrary functions $f : X \to [0,\infty)$?

Answer 1

Because Lebegue's original approach needs them to be measurable in order to be well defined. Then one can show that Lebegue's approach and the one mentioned here are equivalent.

Answer 2

It turns out, that if the integral is finite then the function has to be measurable; see: Royden Fitzpatrick "Real Analysis, 4th edition" chapter 5.3. Their proof works for unbounded functions over infinite-measure spaces too.