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I'm not sure if my attempt is fruitful or not. The exercise is as follows:

Given the function $f: [0,1] \rightarrow \mathbb{R} \cup \{\infty\}$, $~f(x) := \frac{1}{\sqrt(x)}$, $x \neq 0$ and $f(x) := \infty$, $~x = 0$.

Check if f is Lebesgue-integrable.

My assumption is, that $f$ is not integrable (although it is measurable). The reason is, that for $x \rightarrow 0$ the convergence rate of f towards the y-axis is not fast enough.

($\textbf{Question 1:}$ Is there a way to put my very rough and possibly wrong estimation into more mathematical terms?)

Since f(x) $\geq 0$ for each $x\in [0,1]$, I want to show, that there exists a measurable simple function $s, $ $0\le s\le f$, such that sup{$\int_{_{[0,1]}}s ~d\lambda$ : $s$ integrable } $=\infty$.

($\textbf{Question 2:}$ Is it enough to show this?)

Let $I_k := [\frac{1}{k+1},\frac{1}{k}]$ and $s_n := \sum_{k=1}^n \sqrt{k} ~~\chi_{_{I_k}}$. Then for each $n\in \mathbb{N}$ the inequality $0 \leq s_n \leq f(x)$ holds.

To make this a little bit shorter: In the following I would show, that the inequality $\int_{_{[0,1]}}s_{2n} ~d\lambda - \int_{_{[0,1]}}s_n ~d\lambda \geq \frac{1}{2}$ holds. Next I'd conclude, that the growing sequence $\{ \int_{_{[0,1]}}s_n ~d\lambda \}_{n\in \mathbb{N}}$ converges to $\infty$, such that the supremum of this sequence would be $\infty$.

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  • $\begingroup$ This $f$ is Lebesgue integrable. The value at a single point $x=0$ is irrelevant. $\endgroup$ – RRL Nov 14 '19 at 17:50
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Since Lebesgue and Riemann integrals coincide on bounded intervals where the function is Riemann integrable and using the monotone convergence theorem,

$$\int_{(0,1]} x^{-1/2} = \lim_{n \to \infty} \int_{(0,1]}x^{-1/2}\chi_{[1/n,1]}= \lim_{n \to \infty}\int_{1/n}^1 x^{-1/2} \, dx = 2 - \lim_{n \to \infty}\frac{2}{\sqrt{n}} = 2 $$

Along the lines of your attempt, we can also consider the sequence of simple functions,

$$\phi_n(x) = \sum_{k=1}^{2^{n}}(k2^{-n})^{-1/2}\chi_{[(k-1)2^{-n}, k{2^{-n}}]}(x)$$

Again applying the MCT we have

$$\int_{(0,1]}x^{-1/2}=\lim_{n \to \infty}\int_{(0,1]}\phi_n= \lim_{n \to \infty}\sum_{k=1}^{2^{n}}(k2^{-n})^{-1/2}\lambda([(k-1)2^{-n},k{2^{-n}}])\\= \lim_{n \to \infty}\frac1{\sqrt{2^n}}\sum_{k=1}^{2^n}\frac1{\sqrt{k}}= \lim_{n \to \infty}\frac1{\sqrt{n}}\sum_{k=1}^{n}\frac1{\sqrt{k}}=2.$$

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  • $\begingroup$ Recall that MCT states if $f_n$ is a montonically increasing sequence of measurable functions and $\int_A f_n$ is bounded for all $n$, then $f_n$ converges a.e to an integrable function $f$ such that $\int_A f_n \to \int_A f$. $\endgroup$ – RRL Nov 14 '19 at 17:58
  • $\begingroup$ I see my mistake. Naively I was under the impression that the given function behaves similarly to $\frac{1}{x}$ for $x\in (0,1]$, for which the Lebesgue integral does not exist. In a sense, the "rate of asymptotic convergence" of such functions seems to play a role or better: The uniform continuity, which of course has a connection to the existence of the Riemann integral of unbounded functions. Please correct me if I'm wrong. Thanks for your help! $\endgroup$ – JtSpKg Nov 14 '19 at 22:21
  • $\begingroup$ @JtSpKg: You're welcome. Yes there is a connection to improper Riemann integrals of unbounded functions. If a nonnegative function is improperly Riemann integrable then it is Lebesgue integrable. $\endgroup$ – RRL Nov 14 '19 at 23:48
  • $\begingroup$ One more question. Why does the sum run to $2^{-n}$? Shouldn’t it be $2^n$ instead? $\endgroup$ – JtSpKg Nov 15 '19 at 8:47
  • $\begingroup$ Yes it should be $2^n$. Will edit. $\endgroup$ – RRL Nov 15 '19 at 16:32

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