I know its just simple using Riemann Integral. I do not know how can I show the integral of ${x^2}$ over $[0,1]$ equals to $\frac13$ using definition of Lebesgue integral.
Lebesgue Integral of ${x^2}$ over $[0,1]$
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real-analysis
measure-theory
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0Oops, Jonas, you are right. My bad. Thank you for the correction. – 2012-12-17
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In the case of continuous functions, it is known that the Riemann integral and the Lebesgue integral over a finite interval are equal. However, if you don't want to use this result, then what you can do is approximate $f(x) = x^2$ on $[0,1]$ from below by an increasing sequence of simple functions $f_n(x)$, and then calculating the integral of those simple function is just summing the areas of the corresponding rectangles. If you choose $f_n(x)$ is a systematic manner, then it should be possible to evaluate the limit $\lim_{n \rightarrow \infty} \int f_n(x)$, which equals $\int f(x)$ by the monotone convergence theorem.
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0Ya, I know the formula and it seems it comes out right. But I am missing details on proving that sum actually converge to $x^2$. But I can see it is going to converge to $x^2$. – 2013-01-04