Category Theory and Lebesgue Integration.
It's not quite what you were asking, but I feel compelled to link to my own doctoral dissertation here.
http://www.andrew.cmu.edu/user/awodey/students/jackson.pdf
I started from the well known facts that if $\langle \Omega,\mathcal{F}\rangle$ is a measurable space, and $j$ is the countable join topology on $\mathcal{F}$, then:
- The object of measurable real valued functions is the Dedekind real numbers object $\mathbb{R}$ in $\textrm{Sh}_{j}(\mathcal{F})$.
- The presheaf of measures $\mathbb{M}$ is also a sheaf.
- (Lebesgue) integration is definable as a natural transformation $$\textstyle\int\displaystyle:\mathbb{R}^{+}\times\mathbb{M}\to\mathbb{M}.$$
(3) wasn't "well known" as such, but is obvious when you think about it.
The work I did was to find definitions for $\mathbb{M}$ and $\int$ using the internal language of $\textrm{Sh}_{j}(\mathcal{F})$, and to provide categorical proofs of the Monotone Convergence Theorem and the Radon-Nikodym Theorem (spoiler alert: derivatives exist with respect to $\mu$ when the topology of $\mu$-almost everywhere equivalence induces a Boolean topos). These definitions and theorems can then be applied in any topos with a natural numbers object, and not just in the traditional setting of sheaves on a $\sigma$-field.
I don't really understand the question. You give the standard definitions of Lebesgue integration and finally ask if Lebesgue measurable functions form a category? Do you mean if they are the morphisms of a category? Anyway, here is something which might interest you:
One can show that $X=(L^1[0,1],1,\xi)$ is the initial pointed Banach space equipped with a pointed map $\xi : X \oplus X \to X$, see here. Actually we can construct $L^1[0,1]$ this way using abstract nonsense. Applying this to the pointed Banach space $(\mathbb{R},1,m)$ with the mean $m(a,b)=\frac{a+b}{2}$, we obtain a unique map of Banach spaces $\int : L^1[0,1] \to \mathbb{R}$, $f \mapsto \int f(x) \, dx$ such that $\int 1 \, dx =1$ and $$2 \cdot \int f(x) \, dx = \int f\bigl(\tfrac{x}{2}\bigr) \, dx + \int f\bigl(\tfrac{x+1}{2}\bigr) \, dx.$$ I've learned this from a note by Tom Leinster.