For $0In this problem Limit of $L^p$ norm when $p\to0$, the writer states that for $0<p<1$ we have that
$$\Big(\int_\Omega |f|^pd\mu\Big)^{1/p}\leq \frac{1}{R_0} + \Big(\int_\Omega |f_{R_0}|^p\Big)^{1/p} $$
where $\mu$ is a positive measure such that $\mu(\Omega)=1$ and $f_{R_0}=|f|1_A$ where $A={\{x: |f(x)|\geq \frac{1}{R_0}\}}$. I see that
$$\int_\Omega |f|^pd\mu = \int_A |f|^pd\mu + \int_{A^c}|f|^pd\mu \leq \int_\Omega |f_{R_0}|^pd\mu + \frac{1}{R_0^p}$$
But I know that $a^{1/p} + b^{1/p}\leq (a+b)^{1/p}$ where $a,b\geq0$. So I am not sure how to distribute the the $1/p$ power.
 A: It seems to me that the solution for the case $\int_\Omega\log|f|\,d\mu=-\infty$ quoted by the OP is incorrect. For example,  the responder defines $f_R=|f|\mathbb{1}(|f|\geq 1/R)$ and claims that
$$-\log f_R \leq \log(R)$$
This is valid on $\{|f|\geq1/R\}$ but false on $\{|f|<1/R\}$ for there
$$-\log f_R =\infty$$
Also the use of Minkowski's inequality for $0<p<1$ is not valid.

Anyway, if the OP is primarily interested in original problem, i.e. show that
$$\lim_{p\rightarrow0}\Big(\int_\Omega|f|^p\,d\mu\Big)^{1/p}=\exp\Big(\int_\Omega\log|f|\,d\mu\Big)$$
then one may proceed as follows:
For any $a>0$, the function $p\mapsto a^p$ is convex and so the map $\phi_a(p)=\frac{a^p-1}{p}$ is monotone nondecreasing on $(0,\infty)$, and
$\lim_{p\rightarrow0+}\phi(p)=\log(a)$. It follows that if
$$g_p(\omega):=\Big(|f(\omega)|-1\Big) -  \frac{|f(\omega)|^p -1}{p},$$
then $0\leq g_{p'}\leq g_p$ for all $0<p<p'\leq1$, and $g_p\xrightarrow{p\rightarrow0+}|f|-1-\log(|f|)$. An application of  monotone convergence gives
\begin{align}
\lim_{p\rightarrow0+}\Big(\int_\Omega(|f|-1)\,d\mu-\int_\Omega\frac{|f|^p-1}{p}\,d\mu\Big)&=\lim_{p\rightarrow0+}\int_\Omega g_p\,d\mu\\
&=\int_\Omega\lim_{p\rightarrow0+}g_p\,d\mu=\int_\Omega(|f|-1)-\log|f|\,d\mu
\end{align}
whence we conclude that
$$\lim_{p\rightarrow0+}\int_\Omega\frac{|f|^p-1}{p}\,d\mu=\int_\Omega\log|f|\,d\mu$$
regardless of whether $\int_\Omega\log|f|\,d\mu=-\infty$ or $\log|f|\in\mathcal{L}_1(\mu)$.
The conclusion follows from this along with the  inequalities
\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}=\frac{\int_\Omega|f|^p\,d\mu -1}{p}\xrightarrow{p\rightarrow0+}\int_\Omega\log|f|\,d\mu
\end{align}

A short proof that $\phi_a(p)=\frac{a^p-1}{p}$ is monotone nondecreasing in $(0,\infty)$
Claim: If $\phi$ is convex in $(a,b)$, then for any $a<x<u<y<b$
\begin{align}
\frac{\phi(u)-\phi(x)}{u-x}\leq \frac{\phi(y)-\phi(x)}{y-x}\leq \frac{\phi(y)-\phi(u)}{y-u}\tag{1}\label{one}
\end{align}
Indeed, if $u=(1-\lambda)x+\lambda y$, then $\lambda=\frac{u-x}{y-x}$ and from
$$\phi((1-\lambda)x+\lambda y)\leq (1-\lambda)\phi(x)+\lambda\phi(x)$$
we get
$$
\frac{\phi(u)-\phi(x)}{u-x}\leq \frac{\phi(y)-\phi(x)}{y-x}
$$
A similar argument gives the second inequality in \eqref{one}. The converse holds to, that is if \eqref{one} holds for any $a<x<u<y<b$, then $\phi$ is convex in $(a,b)$.
In our case, let $\phi(p)=a^p$ ($a>0$). This is a convex function and
with $x=0$ and $0<u<y$
$$
\frac{\phi(u)-\phi(0)}{u-0}=\frac{a^u-1}{u}\leq \frac{\phi(y)-\phi(0)}{y-0}=\frac{a^y-1}{y}
$$
