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I'm trying to figure out how to compute a particular integral using Lebesgue integration.

For a number $a$, define $f(x) = x^a$ for $0 \lt x \leq 1$, and $f(0) = 0$. Compute $\int_0^1 f$.

Here is what I have so far:
$$\int_0^1 f = \int_{(0,1]}f + 0 = (x^a)*m((0,1]) = x^a$$

I'm not sure if I'm doing this correct. I'd appreciate some help, thanks in advance. This problem appears in Real Analysis by Royden (4th Edition) on p. 84, Exercise 19.

user1234
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    youre not doing it right. the riemann and lebesgue integrals agree (away from zero). instead of taking an improper riemann integral, you can use monotone convergence or something. in any case, you will be using the fundamental theorem of calculus. you do know that $\int x^a=x^{a+1}/(a+1)$ or $\log x$, correct? – yoyo Mar 16 '11 at 17:14
  • Yes, I do know that. But then what does f(0) = 0 have to do with computing this integral? –  Mar 16 '11 at 17:21
  • The value of $f$ at zero won't have any effect on the integral (since a single point has (Lebesgue) measure 0). I think they put it there so $f$ will be defined at 0 for $a<0$. – Brian Mar 16 '11 at 17:34
  • @yoyo: how was that log x? –  Mar 16 '11 at 17:41
  • @Sanchin: What if $a=-1$ ? – Brian Mar 16 '11 at 18:22
  • Then $\int_0^1$f = $\infty$ –  Mar 16 '11 at 22:26
  • This relates to the question asked here https://math.stackexchange.com/questions/2073028/lebesgue-integration-on-0-1-of-x-alpha – Mihai.Mehe Nov 13 '19 at 05:12

1 Answers1

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Your problem seems to a very basic misunderstanding of the difference between definite integrals and functions.

$\int_{(0,1]} f$ is a specific number, not a function.

You cannot just do $\int_{(0,1]} f = f*m((0,1])$.

The statement that $\int_{(0,1]} f = x^a * m((0,1])$ is meaningless.

Aryabhata
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  • That's a really good point. –  Mar 16 '11 at 17:29
  • I guess since the Riemann and Lebesgue integrals agreed away from zero, it's just a simple computation of a definite integral. –  Mar 16 '11 at 17:32
  • r u saying that the answer is $\frac{x^\alpha}{\alpha+1}$ since $m((0,1])=1$ ? – d13 Apr 29 '13 at 11:18
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    @Fd13: No, there should be no $x$ there. It is a number. Not a function. For $\alpha \neq -1$, you get $\frac{1}{\alpha +1}$. So, yes, your answer is partly correct. – Aryabhata Apr 29 '13 at 16:18
  • thankyou for ur help. i was getting confused with the lebesgue integration. – d13 Apr 29 '13 at 17:24