If $\log(f)$ is integrable then $\int \frac{f^p-1}{p} d\mu \to \int\log(f) d\mu$ as $p \to 0$ If $\log(f)$ is integrable then $\int \frac{f^p-1}{p} d\mu \to \int \log(f) d\mu$ as $p \to 0$ (as $p$ decreases to 0) where $\mu$ is lebesgue measure and f is a non-negative integrable function.
My attempt:
$\frac{f^p -1}{p} \to \log(f)$ pointwise. I tried bounding $\frac{f^p -1}{p}$ by some expression involving $\log(f)$ and then use dominated convergent theorem. Please provide hints, not complete answers.
 A: Presumably you are assuming that $f\in L_p(\mu)$ for all $p\in(0,p_0]$ for some $p_0$. Without loss of generality assume $p_0=1$ (why?).

*

*The assumption that $\log(f)\in L_1(\mu)$ implies that $\mu(f=0)=0$, and $|\log(f)|\in L_1(\mu)$

*The further assumption that $f\in L_1(\mu)$ implies that $\mu(\{f>1\})\leq\int f\,d\mu<\infty$
For any $a>0$, the function $g(p)=a^p$ is convex. This implies that $p\mapsto\frac{g(p)-g(0)}{p}=\frac{a^p-1}{p}$ is increasing in $(0,\infty)$.

*

*On $\{f>1\}$, $\phi_p(\omega):=\frac{f^p(\omega)-1}{p}$ decreases to $\log(f)$ as $p\searrow0$, and
$$0\leq\log(f)\leq \phi_p\leq f-1$$

*On $\{0<f\leq 1\}$, $\phi_p$ decreases to $\log(f)$ as $p\searrow0$, and
$$\log(f)\leq \phi_p\leq f-1 \le0$$
Putting things together, for $0<p<1$,
$$|\phi_p|\leq (f-1)\mathbb{1}_{\{f>1\}} + |\log(f)|\mathbb{1}_{\{0<f\leq 1\}}\in L_1(\mu)$$
One  can now invoke dominated convergence.
A: Hint: Use the convexity of the function $p \mapsto f^p$ (for fixed $f > 0$) to show that $\frac{f^p-1}{p}$ decreases to $\log(f)$ as $p$ decreases to $0$.
