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 $\int_X: L^1(X) \to \mathbb{R}$.