Let, $f(x) = \log(x)$. We have that, $$\vert\vert f \vert\vert_{p} = (\int_{(0,1]}|f(x)|^p ~ d\mu)^{\frac{1}{p}}.$$ By the monotone convergence theorem and that the Lebesgue integral is equal to the Riemann integral on a compact interval with a bounded function, we can conclude that: $$\int_{(0,1]} |\log(x)|^p ~ d\mu = \lim_{n \rightarrow \infty} \int^{1}_{\frac{1}{n}}|\log(x)|^p ~ dx.$$ Observe that, $$\lim_{u \rightarrow - \infty } |u|^p e^{u} = 0.$$ Hence, there exists a $C > 0$ such that: $$|u|^p e^{u} \le C, ~ ~ \forall u \in [0,-\infty).$$ Let, $x = e^{2u}$, and we obtain that, $\frac{dx}{du} = 2e^{2u}.$ Hence, $$\lim_{n \rightarrow \infty} \int^{1}_{\frac{1}{n}}|\log(x)|^p ~ dx \le K \lim_{n \rightarrow \infty} \int^{0}_{\frac{-ln(n)}{2}}e^{u} ~ du .$$ Where $K$ is some constant greater than zero consisting of $C$ times other constant terms. Thus, $$ \lim_{n \rightarrow \infty} \int^{0}_{\frac{-ln(n)}{2}}e^{u} ~ du < \infty \implies f \in L^{p}((0,1],\mu).$$
Does my proof look good? Thanks.