Lebesgue integrability and measurable functions

Question

Let $f$ be a nonnegative function on the reals. What does the (Lebesgue) measurability of $f$ have to do with the (Lebesgue) integrability of $\int f$? I've spent some time studying the definition at Wikipedia, and I don't see how measurability enters the equation (other than the fact that it is explicitly mentioned at the start). If $f$ is such that every simple function less than $f$ has integral less than some $x\in\Bbb R$, then $\int f\in\Bbb R$ is defined, regardless of the measurability of $f$. What "essential properties" are lost without the assumption of $f$ being measurable?

Full discosure: I've asked this question before, but I put too much tangentially-related text on the page, and so probably lost some viewers.

Answer 1

One obvious problem would be the loss of linearity. Consider a nonmeasurable set $A\subseteq\mathbb[0,1]$. Then its outer measure is different from its inner measure, i.e. $\mu_*(A)<\mu^*(A)<\infty$ and $$1\ne\mu_*(A)+(1-\mu^*(A))=\int_{[0,1]}\chi_A+\int_{[0,1]}(1-\chi_A)\ne\int_{[0,1]}1=1$$ where $\chi_A$ is the characteristic function of $A$.

Answer 2

Have you ever seen this layer-cake picture of the idea behind Lebesgue integral? (Taken from Wikipedia):

The blue construction is Riemann's integral, the red construction is Lebesgue's integral. Let $f=f(x)$ be the function that you are integrating. Consider the approximation of its integral given by the red rectangles. Each rectangle has area $$A_j= y_j \cdot \text{Lenght of }\{ x\in [a, b]\ :\ y_{j-1}< f(x)\le y_j\},$$ where $y_j$ is the vertical position of the upper corners. Measurability of $f$ tells you exactly that the lenght of the set $\{ x\in [a, b]\ :\ y_{j-1}< f(x)\le y_j\}$ is well-defined.