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 $\infty$.

  • $\begingroup$ This $f$ is Lebesgue integrable. The value at a single point $x=0$ is irrelevant. $\endgroup$
    – 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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.