Recently I have been reading the paper *The categorical origins of Lebesgue integration* by Tom Leinster (<https://arxiv.org/pdf/2011.00412.pdf>). In this paper, he said that：



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> For $n \geq 0$, let $E\_{n}$ be the subspace of $L^{p}[0,1], (1\leq p<\infty)$ consisting of the equivalence classes of step functions constant on each of the (open) intervals $\left(\frac{i-1}{2^{n}}, \frac{i}{2^{n}}\right)$, $\left(1 \leq i \leq 2^{n}\right)$. Write $E=\bigcup\_{n \geq 0} E\_{n}$, which is the space of step functions whose points of discontinuity are dyadic rationals. Then $E$ is dense in the set of all step functions on $[0,1]$, which in turn is dense in $L^{p}[0,1]$; so $E$ is dense in $L^{p}[0,1]$. It follows that $L^{p}[0,1]$ is the colimit (direct limit) of the diagram $E\_{0} \hookrightarrow E\_{1} \hookrightarrow \cdots$ in the category $\mathbf{\text{Ban}}$ of Banach spaces with linear contractions as morphisms.
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I know that the category $\mathbf{\text{Ban}}$ is cocomplete, i.e., the colimit of any diagram exists (because it has coproducts and coequalizers). However, I just have no idea **why the density of $E$ in $L^p[0,1]$ implies just that $L^p[0,1]$ is the colimit of the increasing sequence $E\_{0} \hookrightarrow E\_{1} \hookrightarrow \cdots$ in the category $\mathbf{\text{Ban}}$ of Banach spaces with linear contractions**? I thought about this for two days but didn't figure out anything. Can anyone give a convincing or inspiring explanation?


Any help is appreciated.



Direct limit of the sequence $E_{0} \hookrightarrow E_{1} \hookrightarrow \cdots$ in the category of Banach spaces
A colimit is an object $E$ together with morphisms $i\_n:E\_n\to E$ commuting with the inclusions $i\_{n,m}:E\_n\to E\_m$ (i.e., $i\_m\circ i\_{n,m}=i\_n$) such that, for every sequence of morphisms $f\_n:E\_n\to Y$ with $f\_m\circ i\_{n,m}=f\_n$, there is a unique $f:E\to Y$ with $f\_n=f\circ i\_n$ for all $n$.


You have to show that $i\_n:E\_n\hookrightarrow L^p([0,1])$ satisfy this universal property. Given $f\_n:E\_n\to Y$ as above you can define a linear contraction $\varphi: \bigcup\_{n\in\mathbb N} E\_n \to Y$ by $\varphi(x)=f\_n(x)$ for $x\in E\_n$ (which is well-defined) and, because of the density of the union in $L^p([0,1])$, define $f:L^p([0,1])\to Y$ as the unique continuous extension of $\varphi$.


