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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.

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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?2011-03-16
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    Yes, I do know that. But then what does f(0) = 0 have to do with computing this integral?2011-03-16
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    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 $a<0$.2011-03-16
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    @yoyo: how was that log x?2011-03-16
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    @Sanchin: What if $a=-1$ ?2011-03-16
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    Then $\int_0^1$f = $\infty$2011-03-16

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