Prove that $log(x) in L^1((0,1))$.which has been shown in this post.

The first idea is using the integral $int_epsilon^1 log(x) dx$ to approximate it.which may be the definition in classical analysis.I was bit confused for why this limit: $lim_{epsilon to 0} I(epsilon) = int_0^1 log x dx$ holds in Lebesgue integral. Maybe we need to take an absolute value first $int_epsilon^1 |log(x)|dx$ first then using the Monotone convergence theorem for non-negative function, rest of the proof are the similar ,is my interpretation correct?

The second idea also shows in this post:using the fact that $x^{-alpha} in L^1((0,1))$ for $alpha <1$ then near the origin ,we have $x^{alpha}log x to 0$ for all $alpha >0$ which means exist a small neiborhood near origin says $(0,delta)$ such that if $xin (0,delta)$ we have $|x^alog x|le 1/2$ i.e. $|log x|le frac{1}{2}x^{-a}$ since the RHS is integrable hence log is also integrable near origin,is my interpretation correct?