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