Is $L^P(\mu)$ norm continuous with respect to the underlying measure $\mu$?

Question

Is $L^P(\mathbb{R}^d,\mu)$-norm continuous with respect to the underlying measure $\mu$ in the weak topology? Formally, let $\mu_n$ be a sequence of compactly supported Borel probability measures, absolutely continuous w.r.t. Lebesgue measure, on $\Omega \in \mathbb{R}^d$ that converge weakly to $\mu$, can we say $$\|f||_{L^p(\mu_n)} \to ||f||_{L^p(\mu)}\quad \forall f \in \mathcal{F}$$ for $\mathcal{F}:=\{f:\Omega \to R, \; |f|\leq g \;\text{ for }\; g\in L^p \}$?

Answer 1

$\sup \{\|f\|_{L^{p}(\mu)}: |f|\leq g\} $ is same as $\|g\|_{L^{p}(\mu)}$.

$\mu_n(E) \to \mu(E)$ if $E$ is a $\mu-$ continuity set for $\mu$ but not for arbitrary Borel sets $E$. (I will let you write down a counter-example). Let $g=I_E$ where $E$ is a Borel set such that $\mu_n(E)$ does not tend to $\mu (E)$. Then $\sup \{\|f\|_{L^{p}(\mu_n)}: |f|\leq g\} ^{p}=\mu_n(E)$ which does not converge to $\mu (E)=\sup \{\|f\|_{L^{p}(\mu)}: |f|\leq g\} ^{p}$