Prove $\lim \|f\|_p=\|f\|_\infty$ for Borel measure (Domain has infinite measure)

Question

Let $\mu$ denote a Borel measure on $\mathbb{R}^n$. Assume that $f\in L^1_\mu(\mathbb{R}^n)\cap L_\mu^\infty(\mathbb{R}^n)$. Prove that $$\large\lim_{p\to\infty}\|f\|_{L_\mu^p(\mathbb{R}^n)}=\|f\|_{L_\mu^\infty(\mathbb{R}^n)}$$

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My attempt: I know how to prove it for the case of Lebesgue measure, provided the critical assumption that the space $E$ has finite measure. Here, the method doesn't seem to work since $\mathbb{R}^n$ has infinite measure. Perhaps use the fact that $\mathbb{R}^n$ is sigma-finite? I am unsure of the details though.

We can note that $f\in L_\mu^P(\mathbb{R}^n)$ for any $1\leq p\leq\infty$ since $\int|f|^p\,d\mu=\int|f||f|^{p-1}\,d\mu\leq\|f\|_{L_\mu^\infty}^{p-1}\int|f|\,d\mu<\infty$.

Thanks for any help.

Answer 1

Note

$$\begin{split} \|f\|_{L^{p+1}\ \ (\mu)} &= \left(\int_{\mathbb R^n} |f|^{p+1}d\mu\right)^{\frac{1}{p+1}} \\ &= \left(\int_{\mathbb R^n} |f|^p |f|d\mu\right)^{\frac{1}{p} \frac{p}{p+1}} \\ &=\big( \|f \| _{L^p(\nu)}\big)^{\frac{p}{p+1}}, \end{split}$$

where $d\nu = |f| d\mu$. Since $f\in L^1(\mu)$, $\nu(\mathbb R^n)<\infty$ and so using the finite case (here) we have

$$ \lim_{p\to \infty} \|f\|_{L^p(\mu)} = \|f\|_{L^\infty(\nu)}.$$

and it is easy to show $ \|f\|_{L^\infty(\nu)} = \|f\|_{L^\infty(\mu)}$: Pick $c > \| f\|_{L^\infty(\nu)}$ and $c> 0$. Then by definition,

$$ 0= \nu (|f|\ge c) = \int_{|f|\ge c} |f| d\mu \ge c\mu(|f|\ge c).$$

Thus $\mu(|f|\ge c) = 0$ and $c \ge\|f\|_{L^\infty(\mu)}$. Since $c$ is arbitrary we have

$$ \|f\|_{L^\infty(\nu)}\ge \|f\|_{L^\infty(\mu)}.$$

(Note that $\|f\|_{L^\infty(\mu) }= \inf\{ c>0: \mu(|f|\ge c) = 0\}$).