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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 $x\in (0,\delta)$ we have $|x^a\log x|\le 1/2$ 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?

yi li
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    yes, you take $|\log x|$ and then the monotone convergence theorem, that's correct. So you need to find an increasing sequence ${f_n:(0,1]\rightarrow\mathbb R}$ of positive measurable functions that converge pointwise to $|\log x|$ (hint: find a sequence of increasing sets $B_0\subseteq B_1\subseteq\cdots\subseteq B_n\subseteq\cdots$ and define $f_n=\chi_{B_n}|\log x|$, where $\chi_{B_n}$ is the characteristic function for $B_n$) – Alessandro Apr 06 '21 at 07:39

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  1. Yes, the correct approach is to use Beppo Levi as $\varepsilon\searrow 0$ on $$\int \left\lvert \log (x)1_{(\varepsilon,1]}(x)\right\rvert\,dx=\int_\varepsilon^1-\log(x)\,dx=\varepsilon-\varepsilon\log\varepsilon$$ The important thing is that you must integrate the absolute value, since there is always the possibility that $\lim_{x\to 0}\int_x^1 f(x)\,dx$ exists in $\Bbb R$ but $\lim_{x\to0}\int_x^1\lvert f(x)\rvert\,dx=\infty$. Notice, however, that if you can divide the domain of a, say, locally Riemann integrable function into a finite interval where the function is bounded and an interval where the function doesn't change sign (such is the case for $\log$ on $(0,1]$), then this obstruction cannot occurr, and therefore you can make do with convergence of the improper Riemann integral of the function.

  2. True, but it seems to be more of the same as above, just with different functions.

  • thank you do you mean for the bounded part if oscillation occurs on the compact subset the integral $|\int_K fdx| \le C m(K) <\infty$ so it does not change the property of integral?While the part that does not change the sign,we may take $\lim \int_{K_2/B_\epsilon} \text{sgn}(f) f dx = \int_{K_2} |f| dx <\infty$ hence if locally Riemann integrable is also Locally Lebesgue integrable? – yi li Apr 06 '21 at 08:15