# Can $L^1_{loc}$ be represented as colimit?

Let $$L^1_{loc}$$ denote the set of all functions from $$\mathbb{R}$$ to itself which are locally integrable. For every infinite compact subset $$K\subseteq \mathbb{R}$$, let $$L^1_{m_K}$$ denote the space of Lebesgue measurable functions supported on $$K$$.

Clearly the collection $$\mathcal{K}$$ of all such compact subsets of $$\mathbb{R}$$ form a poset wrt inclusion $$i^{K_1}_{K_2}:K_1\hookrightarrow K_2$$ if and only if $$K_1\subseteq K_2$$, for $$K_i \in \mathcal{K}$$. Therefore, we may define the colimit $$\operatorname{colim}_{\mathcal{K}} L^1_{m_K},$$ in Top.

How are $$\operatorname{colim}_{\mathcal{K}} L^1_{m_K}$$ and $$L^1_{loc}$$ related?

Note/Edit: Top is the category of topological spaces and continuous maps and LCS is the category of locally convex spaces and continuous linear maps.

Related: $L^1_{\mu}$ as limit

• Using the standard definition of L^1, a function in L^1_{m_K} can take arbitrary values outside of K. Thus L^1_{m_K1} is not a subset of L^1_{m_K2} for K1⊂K2. – Dmitri Pavlov Nov 30 '19 at 23:48
• If you change the definition of L^1_{m_K} to say that it is a subspace of L^1 consisting of functions supported on K, the resulting colimit is isomorphic to L^1_loc, pretty much by definition of L^1_loc. – Dmitri Pavlov Dec 1 '19 at 0:13
• @DmitriPavlov If $L^1_{m_K}$ is the space of $L^1$-functions with support in $K$ then we have an inductive spectrum of spaces and hence an inductive limit (which is the same as a colimit). The corresponding topology in TOP (and also in the category LCS of locally convex spaces) is much finer than the Frechet topology of $L^1_{loc}$ (which is the reverse or projective limit in LCS with respect to the restriction mappings). – Jochen Wengenroth Dec 1 '19 at 10:30
• @JochenWengenroth I would expect $\operatorname{colim} L^1_{m_K}$ (essentially definition almost) to be at-least as fine as $L^1_{loc}$, but why ("how much") finer? From your argument I see that it should be finer, but is there a concrete example of [functions] converging in $L^1_{loc}$ but not in $\operatorname{colim}L^1_{m_K}$, for example? – AIM_BLB Dec 1 '19 at 11:57
• @JochenWengenroth: The original question (before it was edited) clearly talked about a set L^1_loc, so the colimit was assumed to be in the category of sets. Now that the question was retroactively edited, my answer no longer makes sense. – Dmitri Pavlov Dec 1 '19 at 15:39

Interpreting the question as Dmitri Pavlov I assume that $$L^1_K$$ is the space of $$L^1$$-functions with support in $$K$$ so that we have an inductive spectrum of Banach spaces (identifying almost everywhere equal functions). The colimit (=inductive limit) in the category of locally convex spaces (i.e., the union endowed with the finest locally convex topology making the inclusions from all $$L^1_K$$ continuous) is then a complete locally convex space (by a classical result of Dieudonne and Schwartz about strict countable inductive limits -- instead of all compact sets it is of course enough to consider a countable exhaustion). On the other hand it is dense in $$L^1_{loc}$$ (the projective limit of all $$L^1_K$$ with respect to the restrictions). Therefore, the inductive limit topology is strictly finer. (There are several other ways to see this. For example, a countable inductive limit of normed spaces $$X_n$$ such that $$X_n\neq X_{n+1}$$ is never metrizable.)
If you consider uncountable colimits in LCS the situation is slightly different: As a Frechet spaces $$L^1_{loc}$$ is ultrabornological and hence the inductive limit of Banach spaces, namely of all Banach spaces generated by absolutely convex closed bounded sets (generated means that you take the linear span endowed with the Minkowski functional). You can describe these spaces as weighted $$L^1$$-spaces $$\{f\in L^1: \int |f|wd\mu<\infty\}$$ with suitable weights.