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.


1 Answer 1


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 $L^1$.


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