Prove that $ \lim_{p\rightarrow 0^+}(\frac{1}{|E|}\int_E f^p)^{\frac{1}{p}}=\exp(\frac{1}{|E|}\int_E \log f) $

Question

I ask this question two days ago, but no one answer or comment. Thus, I reedit my question and edit more about mine thought. Hope someone can help me prove that. Here the question:

Let $E$ be a measurable set in $\mathbb{R^n}$ with $|E| <\infty$. Suppose that $f > 0$ a.e. in $E$ and $f, \log f \in L^1(E)$. Prove that $$ \lim_{p\rightarrow 0^+}\left(\frac{1}{|E|}\int_E f^p\right)^{\frac{1}{p}}=\exp\left(\frac{1}{|E|}\int_E \log f\right) $$

First, I think to show that $$\frac{f^p-1}{p} \rightarrow \log f$$ Whlie I don't know what should I do next that can help to get the result can I take $\frac{1}{p}$ down by $\log$? or first step I think should be change. I think to use LDCT, but I can't find $\phi$ to control $f^p.$

---

Another question is let $f$ be measurable, nonnegative, and finite a.e. in a set $E$. Prove that for any nonnegative constant $c$, $$\int_E e^{cf(x)}dx=|E|+c\int_0^\infty e^{c\alpha}\omega_f(\alpha)d(\alpha)$$

I think that deduce that $e^{cf} \in L(E)$ if $|E| < \infty$ and there exist constants $C_1$ and $c_1$ such that $c_1>c$ and $\omega_f(\alpha) \leq C_1e^{-c_1\alpha}$ for all $\alpha>0$. I try to use $$\int_E \phi(f)=-\int_{-\infty}^{\infty}\phi(\alpha)d\omega(\alpha)$$ Thus, $$\int_E e^{cf}=-\int_{-\infty}^{\infty}e^{c\alpha} d\omega(\alpha)=-\int_{-\infty}^{0}e^{c\alpha} d\omega(\alpha)-\int_{0}^{\infty}e^{c\alpha} d\omega(\alpha)$$ What can I do next to get the result?

How can I prove this two questions?

Answer 1

The first question has already been answered in MSE, see here for example.

---

As for the second question, assuming that $f\geq0$ and $\mu(E) <\infty$, one can use Fubini's theorem to get

\begin{align} \int_Ee^{cf}d\mu&=\int_E\big(e^{cf}-1+1\big)d\mu=\mu(E)+\int_E(e^{cf}-1)\,d\mu\\ &=\mu(E)+ c \int_E\Big(\int^f_0 e^{ct}\,dt\Big)\,d\mu\\ &\stackrel{Fubini}{=}\mu(E)+ c\int^\infty_0\Big(\int_E e^{ct}\mathbb{1}_{\{f >t\}}\,d\mu\Big)\,dt\\ &=\mu(E)+ c\int^\infty_0e^{ct}\mu(f>t)\,dt \end{align}

Comment:

There are some treatments of integration theory in which the Lebegue integral of a measurable nonnegative function $f$ is defined as $$\int_E f\,d\mu:=\int^\infty_0\mu(x\in E: f(x)>t)\,dt$$ where the integral in the right-hand-side is taken in the sense of improper Riemann integral. See Lieb, E. and Loss, M., Analysis, AMS, 2001 for example. In such case, an alternative solution can be obtained along the following lines:

\begin{align} \int_E e^{cf}\,d\mu&=\int^\infty_0\mu(e^{cf}>t)\,dt=\int^\infty_0\mu(f>\frac1c\log t)\,dt\\ &\stackrel{u=\frac1c\log t}{=}c\int^\infty_{-\infty}\mu(f>u)e^{cu}\,du\\ &=c\int^0_{-\infty}\mu(f>u)e^{cu}\,du+ c\int^\infty_0\mu(f>u)e^{cu}\,du\\ &=\mu(E)c\int^0_{-\infty}e^{cu}\,du+ c\int^\infty_0\mu(f>u)e^{cu}\,du\\ &=\mu(E)+ c\int^\infty_0\mu(f>u)e^{cu}\,du \end{align}