Lemma 1: Assume $ f \in L^1 $. Then $ \lim_{p \rightarrow 0} \int_X |f|^p d\mu = \mu (\left\{x \in X|f(x) \neq 0 \right\}) $
Proof: We have $ \lim_{p \rightarrow 0} \int_X |f|^p d\mu = \lim_{p \rightarrow 0} \int\limits_{\left\{|f|>1\right\}} |f|^p d\mu + \lim_{p \rightarrow 0} \int\limits_{\left\{0<|f| \leq 1\right\}} |f|^p d\mu $.
Now $ \lim_{p \rightarrow 0} \int\limits_{\left\{0 < |f| \leq 1\right\}} |f|^p d\mu = \mu (\left\{x \in X|0<f(x) \leq 1 \right\}) $ by Bounded Convergence Theorem, since $ 0 \leq |f|^p \leq 1 \in L^1 $ when $ |f| \leq 1 $, and $ \mu(X) < \infty $
Similarly, $ \lim_{p \rightarrow 0} \int\limits_{\left\{|f|>1\right\}} |f|^p d\mu = \mu (\left\{x \in X|f(x) > 1 \right\}) $ by MCT, since we're bounded above by $ |f| \in L^1 $, for $ p<1 $. Adding yields the result. $ \square $
Lemma 2: If $ f \in L^1(X), f>0 $ a.e., $ \psi(t) := \int_X |f|^td\mu $ is differentiable $ \forall \ t \in (0,\frac{1}{2}) $, and $ \psi'(t) = \int_X |f|^t \log|f| d\mu $.
Proof: $ \lim_{h \rightarrow 0} \frac{f(t+h)-f(t)}{h} = \lim_{h \rightarrow 0} \int_X \frac{|f|^{t+h}-|f|^t}{h} d\mu = \lim_{h \rightarrow 0} \int_X |f|^\xi \log|f|d\mu $, for some $ \xi\in(t,t+h) $, by the Mean Value Theorem. For $ |f|>1, |f| > |f|^{2\xi} > |f|^\xi(|f|^\xi-1) \geq |f|^\xi log|f| \downarrow |f|^tlog|f| >0 $, and for $ 0<|f|\leq1, 0 \geq |f|^\xi \log|f| \downarrow |f|^tlog|f| $, so two applications of MCT allow us to pass the limit inside the integral, yielding the result. $ \square $.
Corollary 3: Under the assumptions of Lemma 2, $ \varphi(t) := \log(\psi(t)) $ is differentiable $ \forall t \in (0,1) $, and $ \varphi'(t) = ||f||_t^{-t}\int_X |f|^t \log|f| d\mu $
Proof: Chain rule.
Main Result: Assume $ f>0 $ a.e.
$ \lim_{p \rightarrow 0} \log||f||_p = \lim_{p \rightarrow 0} \frac{\varphi(p)}{p} = \lim_{p \rightarrow 0} \frac{\varphi(p)-\varphi(0)}{p} $, since $ |f|>0 a.e. \Rightarrow |f|^0 = 1 \ a.e. \Rightarrow \varphi(0)=\log(\int_X d\mu) = 0 $, since $ \mu(X)=1 $.
By the Mean Value Theorem, $ \lim_{p \rightarrow 0} \frac{\varphi(p)}{p} = \lim_{p\rightarrow 0}\varphi'(q) = \lim_{p\rightarrow 0}||f||_q^{-q}\int_X |f|^q \log|f| d\mu $, for some $ q \in (0,p) $. As $ p \rightarrow 0, q\rightarrow 0 $, so we can take the limit on the right.
We have $ \lim_{q \rightarrow 0} \int_{f \leq 1} |f|^q \log|f| d\mu = \int_{f \leq 1} \log|f| d\mu $ by MCT, since $ 0 > |f|^q \log|f| \downarrow log|f| $, and $ \lim_{q \rightarrow 0} \int_{|f| > 1} |f|^q \log|f| d\mu = \int_{f > 1} \log|f| d\mu $ by DCT, since $ |f|>e \Rightarrow |f|^q \log|f| < (|f|\log|f|)^q < (|f|(|f|-1))^q < |f|^{2q} < |f| \in L^1 $ if $ q < \frac{1}{2} $ since $ \log|f| \leq |f|-1. $. If $ 1<|f|<e $ convergence follows from the Bounded Convergence Theorem. Summing yields the result, since we showed $ \int_{f > 1} \log|f| d\mu < \infty $
By Lemma 1, $ \lim_{p \rightarrow 0} ||f||_p^{-p} = 1 $.
Combining these yields $ \lim_{p \rightarrow 0} \log||f||_p = \int_X \log|f| d\mu $. Finally,
$ \exp(\int_X \log|f| d\mu) = \exp (\lim_{p \rightarrow 0} \log||f||_p)= \lim_{p \rightarrow 0} ||f||_p $ by continuity of the exponential function. This proves the statement if $ |f|>0 $ a.e.
General Case: If f = 0 a.e. the statement is trivially true, using the convention $ \log(0) = -\infty, e^{-\infty} = 0 $.
Now let A denote the set where $ |f| = 0, 0 \leq \mu(A) < 1 $. We define a new measure $ \nu = \frac{\mu}{\mu(A^c)} $. Under this measure and restricting our attention to $ A^c $, $ A^c $ is a probability space, $ |f|>0 $ a.e. on $ A^c $, so we reduce to the special case considered above. As mentioned before, the statement is true trivially on A. This concludes the proof. $ \square $
