Integrate and measure problem.

Solution 1:

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.

Solution 2:

$1$. 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 $$ \lim_{p\to0^+}\|f\|_p=e^{\int_{X}\log|f(x)|\,\mathrm{d}x}\tag{2g} $$