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 $K_i$.