Talk:Locally integrable function

Euclidean space

I don't know much about this subject, which is what led me to this page. But the article gives a weaker definition than is apparently available, saying the more abstract setting is not used very often. Here is at least one reader who needed it! Maybe somebody who knows more about this subject than I do can improve the article. Preceding unsigned comment by 216.165.151.43

In this vein, here are some useful references:

• Schilling & Kuhn (2021) and Driver (2012) define ${\mathcal {L}}_{\mathrm {loc} }^{1}(\mu )$ given an arbitrary topological space $(X,{\mathcal {T}})$ and an arbitrary measure $\mu$ on the Borel $\sigma$-algebra $\sigma ({\mathcal {T}})$
• Edwards & Gaudry (1977) and Filter & Weber (1997) define ${\mathcal {L}}_{\mathrm {loc} }^{1}(\mu )$ given an arbitrary measure space $(X,{\mathcal {A}},\mu ):$ a function $f:X\to {\bar {\mathbb {R} }}$ is in ${\mathcal {L}}_{\mathrm {loc} }^{1}(\mu )$ iff for every $A\in \mu ^{\mathrm {pre} }\mathbb {R} ,$ (i.e., $A$ is measurable and has finite measure), $fe_{A}^{X}\in L^{1}(\mu _{A}),$ where $\mu _{A}$ is the subspace measure induced on $A$. [ Jacobs (1978) gives the definition for an arbitrary "measure"/"$\sigma$-content" on an arbitrary "local $\sigma$-ring" (a.k.a. $\delta$-ring).] [Filter & Weber give a version for an arbitrary ring of sets ${\mathcal {A}}$. At first glance, Filter & Weber may appear to say that any locally integrable function will be integrable; taking $A=X$ in their definition yields $f\in L^{1}(\mu )$. However, their definition corresponds to our $\sigma$-algebra definition only when we consider the ring $\mu ^{\mathrm {pre} }\mathbb {R}$ of finite-measure measurable sets. In this case, it is obvious that their definition coincides with ours.]

Thatsme314 (talk) 10:16, 9 May 2022 (UTC)

Examples

In what sense is the final example a solution to the given differential equation on all of $\mathbb {R}$? I get that the restriction of this function to ${\mathbb {R} }-\{0\}$ does not extend to a distribution on $\mathbb {R}$, but it's unclear how the function given (with $f(0)=0$) can be interpreted as some kind of weak solution of the given equation across zero, since it's not integrable. Spireguy (talk) 15:05, 21 May 2023 (UTC)