This appears as Ex. $3.5(d)$ in Rudin's Real and Complex Analysis.
Suppose $f:X\to\mathbb C$ is a measurable function on $X$, $\mu$ is a positive measure on $X$, and $\mu(X) = 1$. Assume that $\|f\|_r < \infty$ for some $r >0$, and prove that $$\lim_{p\to 0} \|f\|_p = \exp\left\{\int_X \log |f|\ d\mu \right\}$$ if $\exp\{-\infty\}$ is defined to be $0$.
I saw a related proof here, and I write this post to confirm the correctness of my proof which fills in all details and removes apparently unnecessary assumptions from the linked answer.
My work: For $0 < p < \infty$, we have $$\tag{1}\int_X \log|f|\ d\mu \le \log\|f\|_p$$ using Jensen's inequality (and that $\log$ is a concave function).
From Q5(a) of Rudin, we have $$\tag{2} 0 < r < s \le \infty \implies \|f\|_r \le \|f\|_s$$
From the graph of $x \mapsto \frac{a^x - 1}{x}$, $\log a \le \frac{a^x - 1}{x}$ for $x > 0$. Put $a = \left(\int_X |f|^{1/n} \ d\mu\right)^{n}$ to get $$\tag{3} \log \|f\|_{1/n} \le \int_X \frac{|f|^{1/n} - 1}{1/n}\ d\mu$$ Let $A = \{x: |f(x)| \ge 1\}$ and $B = \{x: |f(x)| < 1\}$. $$\int_X \frac{|f|^{1/n} - 1}{1/n}\ d\mu = \int_A \frac{|f|^{1/n} - 1}{1/n}\ d\mu + \int_B \frac{|f|^{1/n} - 1}{1/n}\ d\mu$$ Recall that $\|f\|_r < \infty$ for some $r > 0$. For $a >0$, $x \mapsto \frac{a^x - 1}{x}$ is an increasing function in $x$. $\frac{|f|^{1/n} - 1}{1/n}$ is dominated by $\frac{|f|^{r} - 1}{r}$ for large enough $N$, where the latter is integrable. On $B$, $\frac{|f|^{1/n} - 1}{1/n} < 0$, so $\frac{1 - |f|^{1/n}}{1/n} > 0$. In fact, on $B$, as $n\to\infty$, $\frac{1 - |f|^{1/n}}{1/n}\nearrow -\log|f|$ pointwise. We can apply MCT on $B$, and the DCT on $A$, to get $$\lim_{n\to\infty}\int_X \frac{|f|^{1/n} - 1}{1/n}\ d\mu = \lim_{n\to\infty} \int_A \frac{|f|^{1/n} - 1}{1/n}\ d\mu - \lim_{n\to\infty}\int_B \frac{1 - |f|^{1/n}}{1/n}\ d\mu\\ = \int_A \log|f|\ d\mu + \int_B \log|f|\ d\mu= \int_X \log|f|\ d\mu$$
Due to inequality (2), and inequality (1), $\lim_{p\to 0} \log \|f\|_p$ exists.
In inequality (3), take limits as $n\to\infty$ to get $$\lim_{n\to\infty}\log\|f\|_{1/n} \le \int_X \log|f|\ d\mu$$ $$\lim_{p\to 0}\log\|f\|_{p} \le \int_X \log|f|\ d\mu \tag{4}$$ Taking $p\to 0$ in inequality (1), we have $$\lim_{p\to 0}\log\|f\|_{p} \ge \int_X \log|f|\ d\mu \tag{5}$$ $$\implies \lim_{p\to 0}\log\|f\|_{p} = \int_X \log|f|\ d\mu \tag{6}$$ Since $\log$ is continuous, $$\lim_{p\to 0} \|f\|_p = \exp\left\{\int_X \log |f|\ d\mu \right\}$$
Thank you! Please help fill in the gaps if any, and in particular, check if the MCT/DCT part is good.
