0
$\begingroup$

I am making a report for a math course I am taking. I need to investigate whether $x^a$ is Lebesgue-integrable. My plan was to do it this way: I know from the course that Riemann integrability implies Lebesgue integrability and that their values should be the same.

I know from my Calculus course that $x^a$ is Riemann-integrable if $a>-1$ with value $\dfrac 1 {1+a}$. Now I want to use this to prove that the Lebesgue integral has the same value, but for that I need to know whether improper Riemann implies Lebesgue integrability. Is this true, and how can you prove it? If not, how could I investigate for which values of $a$, $x^a$ belongs to the $L^p$-space?

  • 0
    It would help if you reminded the readers of "type 2" means, since the US terminology is not necessarily used across the whole globe. – 2017-01-12
  • 0
    ah yes, type 2 means that you integrate over a bounded [a,b] and that the function might be unbounded at some point in the interval. So the task also was to integrate over [0,1) – 2017-01-13
  • 0
    Given that $\frac 1 {x^a}$ is not defined in $0$ when $a>0$, I believe that your interval is in fact $(0,1]$, not $[0,1)$. Right? – 2017-01-13
  • 0
    My professor did write [0,1), but I already doubted whether that was a spelling error or that was his intention. Anyway, a lot of thanks for the answer, I think this is what I am going to use. – 2017-01-16

2 Answers 2

0

I suppose you are integrating over $[1,\infty)$. Let $f(x) = x^a$ and suppose that $a>-1$. Let $f_n = f \chi_{[1,n]}$ and use Lebesgue's monotone convergence theorem.

To answer the question in the title, no it doesn't. Consider $f(x) = \frac{\sin x}{x}$. This function is Riemann integrable but not Lebesgue integrable.

  • 0
    Thanks for the counter example, However, I forgot to say I should integrate over [0,1). – 2017-01-13
0

To clarify, it seems that you are asking when the function $\dfrac 1 {x^a}$ is in $L^p ((0,1])$ for some given $p \ge 1$. To answer this, the most straightforward approach is to use Levi's monotone convergence theorem.

Let $f: (0,1] \to \Bbb R$ be $f(x) = \dfrac 1 {x^{pa}}$. Let $f_n : (0,1] \to \Bbb R$ be $f_n (x) = \begin{cases} 0, & x< \frac 1 n \\ f(x), & x \ge \frac 1 n \end{cases}$. Notice that $f_n \to f$ and that $f_n$ is integrable. Notice also that $f_n \le f_{n+1}$. Levi's theorem guarantees then that

$$\int \limits _{(0,1]} f(x) \ \Bbb d x = \int \limits _{(0,1]} \lim _{n \to \infty} f_n (x) \ \Bbb d x = \lim _{n \to \infty} \int \limits _{(0,1]} f_n (x) \ \Bbb d x = \lim _{n \to \infty} \int \limits _0 ^1 f_n (x) \ \Bbb d x = \lim _{n \to \infty} \int \limits _{\frac 1 n} ^1 \frac 1 {x^{pa}} \ \Bbb d x = \\ \lim _{n \to \infty} \begin{cases} \log n, & a = \frac 1 p \\ \frac 1 {1-pa} ( 1 -n ^{-1+pa} ), & a \ne \frac 1 p \end{cases} = \begin{cases} \infty, & a \ge \frac 1 p \\ \frac 1 {1-pa}, & a < \frac 1 p \end{cases} ,$$

which means that $f \in L^p ((0,1]) \iff a < \dfrac 1 p$.