Let $(E,\mathcal E,\lambda)$ be a measure space, $p$ be a probability density on $(E,\mathcal E,\lambda)$, $\mu:=p\lambda$, $$\iota g:=\left(E\ni x\mapsto\begin{cases}\displaystyle\frac{g(x)}{p(x)}&\text{, if }p(x)>0\\0&\text{, otherwise}\end{cases}\right)\;\;\;\text{for }g:E\to\mathbb R,$$ $r\ge1$, $$\mathcal L^r:=\left\{g:E\to\mathbb R:g\text{ is }\mathcal E\text{-measurable with }\{p=0\}\subseteq\{g=0\}\text{ and }\int_{\{\:p\:>\:0\:\}}\frac{|g|^r}{p^{r-1}}\:{\rm d}\lambda<\infty\right\}$$ and $$K_r:=\left\{g\in\mathcal L^r:g\ge0,g\text{ is bounded},\int g\:{\rm d}\lambda=0\text{ and }\int_{\{\:p\:>\:0\:\}}\frac{g^r}{p^{r-1}}\:{\rm d}\lambda\le1\right\}.$$
How can we show that $\iota K_r$ is a closed subset of $L^r(\mu)$?
Maybe the easiest way is to note that $\mathcal L^r$ equipped with $$\left\|g\right\|_{L^r}^r:=\int_{\{\:p\:>\:0\:\}}\frac{|g|^r}{p^{r-1}}\:{\rm d}\lambda\;\;\;\text{for }g\in\mathcal L^r$$ is a semi-normed $\mathbb R$-vector space ($\left\|g\right\|_{L^r}=0$ if and only if $g=0$ $\lambda$-almost everywhere) and $\iota$ is a linear isometry from $\mathcal L^r$ to $\mathcal L^r(\mu)$. Then it should be sufficient to show that $K_r$ is closed.
Basically you look at a sequence $\{g_n\}$ in $K_r$. Suppose $g_n \to g$ in $L^r$.
Then we have $\int_{p>0} \frac{g^r}{p^{r-1}} d \lambda = \lim\inf\int_{p>0} \frac{g_n^r}{p^{r-1}} d \lambda < \infty$ by Fatou's lemma. We used the fact that $g_n \in L^r$ to ensure its less than infinity.
Next we prove $\int g d\lambda = 0$. $g$ is the $L^r$ limit of a sequence of positive functions and is thus positive $\lambda$ a.e. Hence $\int g d\lambda \geq 0$.
Also by Fatou's lemma $\int g dx \leq \inf \int g_n dx = 0$.
Therefore $\int g dx = 0$.
This is enough to prove that $g \in K_r$ i.e. $K_r$ is closed.