Let $\mu$ be a finite measure on the measure space $(\mathbb{R}^d,\Sigma)$, $K\subset \mathbb{R}$ be compact of positive $\mu$-measure, and defined the finite measure $\nu\ll \mu$ on $(\mathbb{R}^d,\Sigma)$ via its Radon-Nikodym derivative as $\frac{d \nu}{d\mu}\triangleq 1_K$ as well as the modified $L^p$ space $$ \begin{aligned} \mathbb{L}^p_{\nu}(\Sigma) &\triangleq \left\{ f \in L_{\mu,loc}^p(\Sigma): \, \int_{\mathbb{R}^d} |f|^p d\nu <\infty \right\} \\&\subseteq \left\{ f \in Mes(\Sigma): \, \int_{\mathbb{R}^d} |f|^p d\nu <\infty \right\} \mapsto L^p_{\nu}(\Sigma)\; \end{aligned} $$ where the last map is the quotient setting $f\sim g $ iff $f=g$ $\nu$-a.e.
Note! : The usual equivalence relation (defining elements of $Mes(\Sigma)$) is defined wrt $\mu$ and not wrt $\nu$.

Edit: Strict Inclusion:

Because of this, tn general, the inclusion is strict; for example if $\mu=$ Lebesgue measure $p=d=1$, $K=[0,1]$, and $g(t)$ is a measurable function of positive quadratic variation then $g 1_{K^c} \not\in \mathbb{L}_{\nu}^1(\Sigma)$ but $g 1_{K^c} \in L_{\nu}^p(\Sigma)$.


Is $\mathbb{L}_{\nu}^p(\Sigma)$ a dense subspace of $L_{\nu}^p(\Sigma)$? and it is a locally-convex (TVS) but not Banach?


If $K_1\subset K_2$ then $L^p_{\nu_1}(\Sigma)\not\subset L^p_{\nu_2}(\Sigma)$ but $\mathbb{L}^p_{\nu_1}(\Sigma)\subseteq \mathbb{L}^p_{\mathbb{R}^d}(\Sigma)$; where $\nu_i$ is the restriction of $\mu$ to $K_i$.

  • $\begingroup$ I don't quite follow your definition of $\nu$. Is it just the restriction of $\mu$ to $K$? $\endgroup$ – Nik Weaver Jan 14 at 14:25
  • $\begingroup$ @NikWeaver precisely, $\nu$ is the restriction of $\mu$ to $K$, defined via the Radon-Nikodym derivative $\endgroup$ – AnnieTheKatsu Jan 14 at 14:35
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    $\begingroup$ Well, then every function in your space equals one in $L^p_\nu(\Sigma)$ a.e. It isn't really research level ... $\endgroup$ – Nik Weaver Jan 14 at 14:48
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    $\begingroup$ By "one in $L^p_\nu(\Sigma)$" I didn't mean it is equal to $1$ a.e., I meant it is equal to something in $L^p_\nu(\Sigma)$ a.e. $\endgroup$ – Nik Weaver Jan 14 at 15:14
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    $\begingroup$ In that case there seems to be one issue: $\mathbb{L}^p_{\nu_1}(\Sigma)\subseteq \mathbb{L}^p_{\nu_2}(\Sigma)$ whereas $L^p_{\nu_1}(\Sigma)$ is not contained in $L^p_{\nu_1}(\Sigma)$ if $K_1\subseteq K_2$ and $\mu(K_2-K_1)>0$; so then $\mathbb{L}_{\nu}^p(\Sigma)\neq L^p_{\nu}(\Sigma)$ in general.. Is your point that $\mathbb{L}_{\nu}^p(\Sigma)\subset L^p_{\nu}(\Sigma)$? $\endgroup$ – AnnieTheKatsu Jan 14 at 15:26

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