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 $\infty$.