# Odd variant of $L^p$ spaces

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 $$g 1_{K^c} \in L_{\nu}^p(\Sigma)$$.

Question:

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

Remark(s):

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 $$K_i$$.

• I don't quite follow your definition of $\nu$. Is it just the restriction of $\mu$ to $K$? – Nik Weaver Jan 14 at 14:25
• @NikWeaver precisely, $\nu$ is the restriction of $\mu$ to $K$, defined via the Radon-Nikodym derivative – AnnieTheKatsu Jan 14 at 14:35
• Well, then every function in your space equals one in $L^p_\nu(\Sigma)$ a.e. It isn't really research level ... – Nik Weaver Jan 14 at 14:48
• 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. – Nik Weaver Jan 14 at 15:14
• 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 $\mathbb{L}_{\nu}^p(\Sigma)\subset L^p_{\nu}(\Sigma)$? – AnnieTheKatsu Jan 14 at 15:26