Integrate and measure problem. If $f \in L^{p_0}(X,M,\,u)$ for some $0<p_0 \le\infty$, then
$$1. \lim_{p\to0}\int_X |f|^p \, d\mu=\mu(\{x \in X \mid f(x) \ne0\}).$$
And if additional assume $\mu(X)=1$,
then I wanna prove that $f \in L^p(X,M,\,u)$ for some $0<p \le p_0$, and the equation below.
$$2. \lim_{p\to0}\|f\|_p=e^{\int_X\log|f|\,d\mu}$$
I want to know how can I conclude those results. First, I'm trying to use integrate over the set on which $0<|f(x)|\le1$ and use MCT. and the set on which $|f(x)|>1$ use LDCT to prove that. But I can't conclude to measure $\mu$ and how can I approach second fact?
 A: For the second equation: first assume that $f$ does not vanish on a set of positive measure. We have
$$\log\|f\|_p = \frac{1}{p} \log(\int_X |f|^p d\mu) .$$
We apply L'hopital's rule to take the limit as $p\rightarrow 0$. Since $|f|^p \log |f|$ is bounded by either a constant or $|f|^{p_0}$ for small $p$, we can differentiate under the integral sign to get
$$\frac{d}{dp}  \log(\int_X |f|^p d\mu) = \frac{\int_X \log |f| * |f|^p d\mu}{\int_X |f|^p d\mu}.$$
Of course the derivative of the denominator $p$ is just 1. Therefore
$$\lim_{p\rightarrow 0} \log\|f\|_p = \lim_{p\rightarrow 0}\frac{\int_X \log |f| * |f|^p d\mu}{\int_X |f|^p d\mu} = \frac{\int_X \log |f|  d\mu}{\int_X 1 d\mu} = \int_X \log |f|  d\mu ,$$
by dominated convergence. It follows that
$$\lim_{p\rightarrow 0} \|f\|_p = e^{\int_X \log |f|  d\mu } .$$
If $f = 0$ on a set $E$ of positive measure, then by H\'older's inequality (with $p_0^* = p_0/(1-p_0)$),
$$\int_X |f|^p d\mu = \int_X \chi_{E^c} |f|^p d\mu \leq \||f|^p\|_{p_0} \|\chi_{E^c}\|_{p_0^*} = (\int_X |f|^{p p_0})^{1/p_0} \mu(E^c)^{1/p_0^*}.$$
Thus,
$$\|f\|_p \leq \|f\|_{p p_0} \mu(E^c)^{1/pp_0^*}.$$
The first term here is bounded for small $p$ and the second tends to 0 as $p\rightarrow 0$ since $\mu(E^c) < 1$. Thus $\|f\|_p \rightarrow 0$, which equals $e^{\int_X \log |f| d\mu}$ if we interpret $e^{-\infty}$ as 0.
A: $1$. Define
$$
\begin{align}
E^>&=\{x\in X:|f(x)|\ge1\}\\
E^<&=\{x\in X:0<|f(x)|<1\}\\
E^=&=\{x\in X:|f(x)|=0\}
\end{align}\tag{1a}
$$
On $E^>$, $|f(x)|^p$ decreases to $1$ as $p$ decreases to $0$; on $E^<$, $|f(x)|^p$ increases to $1$ as $p$ decreases to $0$; and on $E^=$, $|f(x)|=0$ as $p$ decreases to $0$.
Therefore, by monotone convergence on $E^<$ and dominated convergence on $E^>$,
$$
\begin{align}
\lim_{p\to0^+}\int_{E^>}|f(x)|^p\,\mathrm{d}x&=\mu(E^>)\\
\lim_{p\to0^+}\int_{E^<}|f(x)|^p\,\mathrm{d}x&=\mu(E^<)\\
\lim_{p\to0^+}\int_{E^=}|f(x)|^p\,\mathrm{d}x&=0
\end{align}\tag{1b}
$$
Summing these yields
$$
\lim_{p\to0^+}\int_X|f(x)|^p\,\mathrm{d}x=\mu(\{x\in X:|f(x)|\not=0\})\tag{1c}
$$

$2$. Preliminaries

For $p\gt0$ and $t\ge0$, define
  $$
g_p(t)=\frac{t^p-1}{p}\tag{2a}
$$
  Claim: $g_p(t)$ is non-decreasing in both $p$ and $t$.
$g_p(t)$ is non-decreasing in $t$: This follows from
  $$
g_p^\prime(t)=t^{p-1}\ge0\tag{2b}
$$
$g_p(t)$ is non-decreasing in $p$: As Didier commented, this follows from
  $$
g_p(t)=\int_1^tu^{p-1}\,\mathrm{d}u\tag{2c}
$$
  and because $u^{p-1}$ is non-decreasing in $p$ when $u\ge1$ and non-increasing in $p$ when $0\le u\le1$.
Furthermore, L'Hopital says
  $$
\lim_{p\to0^+}g_p(t)=\log(t)\tag{2d}
$$

$\hspace{1pt}$

Jensen's Inequality says that $h(p)=\|f\|_p$ is non-decreasing in $p$.

$\hspace{1pt}$

Consider an $\epsilon$ neighborhood of $-\infty$ to be $(-\infty,-\frac1\epsilon)$ and let $L=\lim\limits_{p\to0^+}\log(h(p))$.
For any $\epsilon>0$, choose $q>0$ so that $\log(h(q))$ is within an $\frac{\epsilon}{2}$ neighborhood of $L$.
Choose $r>0$ so that $g_r(h(q))$ is within an $\epsilon$ neighborhood of $L$.
If $p<\min(q,r)$, then both $\log(h(p))$ and $g_p(h(p))$ will be within an $\epsilon$ neighborhood of $L$. Therefore, 
  $$
\lim_{p\to0^+}\log(h(p))=\lim_{p\to0^+}g_p(h(p))\tag{2e}
$$

Main Result
Define $E=\{x:|f(x)|>1\}$, then the results above yield
$$
\begin{align}
\lim_{p\to0^+}\log\left(\|f\|_p\right)
&=\lim_{p\to0^+}\frac{\|f\|_p^p-1}{p}\\
&=\lim_{p\to0^+}\int_X\frac{|f(x)|^p-1}{p}\,\mathrm{d}x\\
&=\color{#C00000}{\lim_{p\to0^+}\int_{E}\frac{|f(x)|^p-1}{p}\,\mathrm{d}x}
+\color{#00A000}{\lim_{p\to0^+}\int_{X\setminus E}\frac{|f(x)|^p-1}{p}\,\mathrm{d}x}\\
&=\color{#C00000}{\int_{E}\log|f(x)|\,\mathrm{d}x}
+\color{#00A000}{\int_{X\setminus E}\log|f(x)|\,\mathrm{d}x}\\
&=\int_{X}\log|f(x)|\,\mathrm{d}x\tag{2f}
\end{align}
$$
The left limit, in red, is by Dominated Convergence, while the right limit, in green, is by Monotone Convergence. Exponentiate to get
$$
\lim_{p\to0^+}\|f\|_p=e^{\int_{X}\log|f(x)|\,\mathrm{d}x}\tag{2g}
$$
