In [this](https://mathoverflow.net/questions/165405/category-theory-free-areas-of-pure-math-category-theory-loaded-areas-of-app%29%20have%20not%20been%20simplified%20by%20clever%20categorical%20arguments) question of mine in a comment to the accepted answer, [someone](https://mathoverflow.net/users/4362/paul-siegel) remarked:



> 
> There are ways to express even basic things in analysis, such as the
>  spectral theorem or the Lebesgue integral, using the language of
>  categories. But many of the hard theorems in analysis boil down to
>  subtle estimates which (so far!) have not been simplified by clever
>  categorical arguments.
> 
> 
> 


I replied to this requesting a reference where the Lebesgue integral is expressed using categories but unfortunately I didn't get an answer. So I'm posting it as a question (I wouldn't mind for a reference concerning the spectral theorem either) with the additional kind request to also provide an explanation why it is difficult to describe estimates using category theory. 



Expressing the Lebesgue integral using categories + the difficulty of describing estimates in category theory
Sorry to refer to my own work, but I think this answers your question directly: <http://www.maths.ed.ac.uk/~tl/glasgowpssl/>


That link is to a very short note, but I might as well repeat the result here. Let's agree that a "map" of Banach spaces is a map of norm $\leq 1$, and let's also agree that when $X$ and $Y$ are Banach spaces, we equip $X \oplus Y$ with the norm $\| (x, y) \| = (\|x\| + \|y\|)/2$. 


Now let $\mathcal{C}$ be the category of triples $(X, \xi, u)$ where $X$ is a Banach space, $\xi$ is a map $X \oplus X \to X$, and $u$ is an element of the closed unit ball of $X$ such that $\xi(u, u) = u$.


**Theorem**: The initial object of $\mathcal{C}$ is $(L^1[0, 1], \gamma, 1)$ where $1$ is the constant function $1$ and $\gamma$ concatenates two functions then scales the domain by a factor of $1/2$.


Another object of $\mathcal{C}$ is $(\mathbb{R}, \text{mean}, 1)$. The unique map in $\mathcal{C}$ from the initial object to this object is Lebesgue integration, $\int\_0^1: L^1[0, 1] \to \mathbb{R}$.




---


While I'm at it, I'll add another result that isn't in that note (or written up anywhere yet). This characterizes Lebesgue integrability and integration on arbitrary finite measure spaces.


Let $\mathbf{Meas}$ be the category of finite measure spaces and "embeddings" (by which I mean maps that are isomorphisms to their images). Let $\mathbf{Ban}$ be the category of Banach spaces (with maps as above). 


Let $\mathcal{D}$ be the category of pairs $(F, u)$, where $F$ is a functor $\mathbf{Meas} \to \mathbf{Ban}$ and $u$ assigns to each measure space $X = (X, \mu)$ an element $u\_X \in F(X, \mu)$, subject to two laws: first, $\|u\_X\| \leq \mu(X)$, and second, whenever
$$
Y \stackrel{i}{\longrightarrow} X \stackrel{j}{\longleftarrow} Z
$$
in $\mathbf{Meas}$ with $X = iY \sqcup jZ$ (disjoint union) then $(Fi)u\_Y + (Fj)u\_Z = u\_X$.


**Theorem** The initial object of $\mathcal{D}$ is $(L^1, I)$, where $I\_X \in L^1(X)$ is the constant function $1$.


(In the case of this initial object, the equation "$(Fi)u\_Y + (Fj)u\_Z = u\_X$" says that when $X$ is partitioned into subsets $Y$ and $Z$, the indicator function of $Y$ plus the indicator function of $Z$ is the constant function $1$.) 


Another object of $\mathcal{D}$ is $(K, t)$, where $K$ has constant value $\mathbb{R}$ (or $\mathbb{C}$, depending on our choice of ground field) and $t\_X = \mu(X)$ for a measure space $X = (X, \mu)$. The unique map in $\mathcal{D}$ from the initial object to this object is integration. To spell that out a bit more: the maps in $\mathcal{D}$ are natural transformations satisfying the obvious condition, and in this case, the $X$-component of the unique map $(L^1, I) \to (K, t)$ is $\int\_X: L^1(X) \to \mathbb{R}$.


