Given a function f, check, if it is Lebesgue integrable.

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 $$\infty$$.

• This $f$ is Lebesgue integrable. The value at a single point $x=0$ is irrelevant.
– RRL
Nov 14 '19 at 17:50

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

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

• 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 $\int_A f_n \to \int_A f$.
– RRL
Nov 14 '19 at 17:58
• 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! Nov 14 '19 at 22:21
• @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.
– RRL
Nov 14 '19 at 23:48
• One more question. Why does the sum run to $2^{-n}$? Shouldn’t it be $2^n$ instead? Nov 15 '19 at 8:47
• Yes it should be $2^n$. Will edit.
– RRL
Nov 15 '19 at 16:32