# $L^1_{\operatorname{loc}}$ and weak $L^1$

I think in general we cannot say whether $$L^1_{loc}$$ and $$weak$$ $$L^1$$ include each other or not, e.g. $$f(x)=\frac{1}{x}$$ and $$f(x)=1$$ are in $$weak$$ $$L^1(\lambda)$$ and $$L^1_{loc}(\lambda)$$ respectively, but not in the other space (for $$\lambda$$ the Lebesgue measure).

My curiosity/question is if there are some sophisticated measure spaces where an inclusion can be made?

E.g. I would say that in a finite measure space, $$f(x)=1$$ would be in weak $$L^1$$ as well, but a finite measure space is a rather trivial example since we are just making $$L^1_{loc}$$ be the same as $$L^1$$, so we end up with $$L^1_{loc}\subset weak\ L^1$$ and nothing else since $$f(x)=\frac{1}{x}$$ would not necessarily be in $$L^1$$.

At least these are my current thoughts. Let me know if anything I said is wrong.

One thing you should be careful about is the following. The definition of weak $$L^1$$ space does not involve the topology of the space, and in some sense weak $$L^p$$ spaces don't really talk to the topological structure of the space. On the other hand, $$L^p_{loc}$$ does seriously depend on the topology of the space. For example, if you take $$B\subset \mathbb R^d$$ as the open unit ball, and $$\overline B$$ its closure, then clearly $$L^1_{loc}(\overline B)=L^1(\overline B)$$ ("locally" essentially means "on any compact set"if the underlying space is locally compact); but $$L^1_{loc}(B)$$ is much, much larger than $$L^1(B)$$: it also contains functions that grow arbitrarily fast close to the boundary of $$B$$, so in particular it is not contained in weak $$L^1$$.
So, the trivial example you should think of is that of a compact subset of $$\mathbb R^d$$ (or actually, any compact space with at least some nice structure). Finite measure is still not enough to make the inclusion hold. If the underlying space is not compact, then $$L^1_{loc}$$ is in general not $$L^1$$, and is not contained in weak $$L^1$$ either.
On the other hand, I know that whenever you have a measure that is nontrivial and non-pathological enough (I think it's enough that it does not exclusively contain finitely many atoms, and it is sigma-finite), then weak $$L^1$$ strictly contains $$L^1$$. The only cases I can think of where weak $$L^1$$ coincides with $$L^1$$ are essentially trivial examples, like a measure with a finite number of atoms, or a measure which is (almost) nowhere finite.
I honestly don't know if there are nontrivial examples of any of the two inclusions. There could be some interesting, pathological examples where $$L^1_{loc}$$ is so huge that contains all measurable functions, and so contains $$L^1$$, but it seems that for that, any point should admit a neighbourhood which contains only a finite number of atoms, or something like that. In the other direction I am a bit skeptical: I have the suspect that whenever $$L^1_{loc}$$ does not coincide with $$L^1$$ (which looks like a trivial case), functions are allowed to grow arbitrarily fast close to "the boundary" of the space, and it should always be possible to construct functions in $$L^1_{loc}$$ that are not in weak $$L^1$$.