TITLE: Limit of $L^p$ norm when $p\to0$
QUESTION [23 upvotes]: Let ($\Omega$, $\cal{F}$, $\mu$) be a probability space and $f\in L^1(\Omega)$. Prove that
$$\displaystyle\lim_{p\to 0} \left[ \int_{\Omega}|f|^pd\mu \right]^{\frac{1}{p}}=\exp \left[ \int_{\Omega}\log|f| d\mu  \right],$$
where $\exp[-\infty]=0$. To simplify the problem, we may assume $\log|f|\in L^1(\Omega).$

REPLY [4 votes]: @DavideGiraudo: This is a long comment an so I laid it as an answer.

I did not find a direct way to fix the problem for the case where $\log|f|\notin L_1(\mu)$ in the spirit of our solution.
I did however obtained a slightly different solution  that applies to whether $\log|f|$ is integrable or $\int(\log|f|)_-=\infty$. Here is a sketch:

Basically one notices that for any $a>0$, the map $\phi_a(p)=\frac{a^p-1}{p}$ is monotone nondecreasing on $(0,\infty)$ (due to convexity of $p\mapsto a^p$) and so, $g_p:=(|f|-1)-\frac{|f|^p-1}{p}$ is positive and nondecreasing as $p\searrow0$.  Monotone convergence implies that
$$\lim_{p\rightarrow0+}\int g_p=\int \lim_{p\rightarrow0+}g_p=\int(|f|-1-\log|f|)$$
Thus, $\lim_{p\rightarrow0+}\int\frac{|f|^p-1}{p}=\int\log|f|$ regardless of integrability of $\log|f|$. The conclusion follows now from the inequality $\log(a)\leq a-1$ for all $a>0$ and  Jensen's inequality:
\begin{align}
\int_\Omega\log|f|\,d\mu&= \frac{1}{p}\int_\Omega\log(|f|^p)\,d\mu\leq \frac{1}{p}\log\Big(\int_\Omega|f|^p\,d\,\mu\Big)=\log\|f\|_p\\
&\leq \frac{\|f\|^p_p-1}{p}=\int_\Omega\frac{|f|^p -1}{p}\,d\mu\xrightarrow{p\rightarrow0+}\int_\Omega\log|f|\,d\mu
\end{align}