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Consider the vector space $L^p_{\text{left-loc}}$ of measurable functions $f:[0,1]\to\mathbb R$ so that for all $x\in(0,1]$ there exists $\delta>0$ so that $f|_{[x-\delta,x]}\in L^p$. Does this space have a name or have such functions been studied?

This is essentially a local $L^p$ space but integrability is only required "locally to the left". For example, if $0=a_1<a_2<\dots<a_n<a_{n+1}=1$ and $f_k\in L^p_{\text{loc}}((a_k,a_{k+1}])$ (meaning $L^p$ restricted to any compact subset of the half open interval), then $f_1+\dots+f_n$ is in $L^p_{\text{left-loc}}$. There is no control to how badly the function can blow up at a point, as long as it only blows up on the right side of the point.

In fact, as Christian Remling remarked in a comment below, for any function $f\in L^p_{\text{left-loc}}$ there is a unique finite or countable partition of $(0,1]=\bigcup_i(a_i,b_i]$ so that $f|_{(a_i,b_i]}\in L^p_{\text{loc}}\setminus L^p$ — except possibly if $a_i=0$. For any $x\in(0,1]$ there is a maximal half open interval $(a,b]$ containing $x$ so that $f$ is locally in $L^p$ on this interval, and the partition is made of such maximal intervals, the partition is maximal. This gives an alternative characterization of the space.

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  • $\begingroup$ This seems to be all functions $f$ for which there is some partition $(0,1]=\bigcup (a_j,b_j]$ (finite or countable) with $f\in L^p(a_j,b_j)$ for all $j$. (For each $x$, take the maximal half-open interval with $x$ in it on which $f$ is $L^p$.) $\endgroup$
    – Christian Remling
    Jul 22, 2015 at 21:36
  • $\begingroup$ @ChristianRemling, that seems to be an equivalent definition of $L^p_{\text{left-loc}}$, but there need not be a maximal half open interval of integrability. For example $f(x)=x^{-3}$ is in the space but I see no natural way of partitioning $(0,1]$ for it. For any $x\in(0,1)$ the union of all intervals containing $x$ on which $f$ is integrable is $(0,1]$ (this is essentially what the inf and sup do) on which $f$ is not integrable. $\endgroup$
    – Joonas Ilmavirta
    Jul 23, 2015 at 6:42
  • $\begingroup$ Right, what I wrote needs more fixing: given $x$, we want the maximal interval $(a,b]$ with $x$ in it such that $f$ is locally $L^p$ on $(a,b]$. This produces a well defined partition of $(0,1]$, and gives a rephrasing of your definition along the lines of my original comment. $\endgroup$
    – Christian Remling
    Jul 23, 2015 at 17:51
  • $\begingroup$ @ChristianRemling, that seems to work. But unless someone has encountered such spaces, it doesn't quite answer the question. I will consider adding it to the question, though. $\endgroup$
    – Joonas Ilmavirta
    Jul 23, 2015 at 20:29

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