$$\int _0^1 x^2 dx=\frac{1}{3}x^2 \bigg|_{0}^{1}=\frac{1}{3}$$ This solution is the Riemann integral. (strictly speaking, fundamental theorem of calculus)
I want to solve with the Lebesgue integral instead of the Riemann integral.
So,
$$\int _I x^2 d\mu(x), \quad I=[0,1],\quad \mu: Lebesgue\, measure(i.e.\, \mu(I)=1).$$
sol) Let us decompose a given closed interval $I=[0,1]$ of y-axis into n subintervals
$\displaystyle y_1=\frac{1}{n}, y_2=\frac{2}{n}, \dots, y_n=\frac{n}{n}, $
To find each measure, so $\displaystyle f^{-1}(x)=\sqrt x$,
$\displaystyle A_1=f^{-1}\left ([0,\frac{1}{n}]\right), A_2=f^{-1}\left ([\frac{1}{n},\frac{2}{n}]\right), \dots, A_n=f^{-1}\left ([\frac{n-1}{n},\frac{n}{n}]\right)$,
$$\sum_{n=1}^{\infty}y_i \mu(A_i)=\lim_{n \rightarrow \infty}\left\{\bigg(\frac{1}{n}\sqrt\frac{1}{n}\bigg)+\frac{2}{n}\bigg(\sqrt\frac{2}{n}-\sqrt\frac{1}{n}\bigg)+\frac{3}{n}\bigg(\sqrt\frac{3}{n}-\sqrt\frac{2}{n}\bigg)+\cdots+\frac{n}{n}\bigg(\sqrt\frac{n}{n}-\sqrt\frac{n-1}{n}\bigg)\right\}$$
$$=\lim_{n \rightarrow \infty}\left\{\frac{1}{n\sqrt{n}}\bigg(n\sqrt{n}-\sum_{k=1}^{n-1}{\sqrt{k}}\bigg)\right\}$$ $$=\lim_{n \rightarrow \infty} \left\{ 1-\frac{\sum_{k=1}^{n-1}{\sqrt{k}}}{n\sqrt n}\right\}=\frac{1}{3} \qquad (\because)\, Taylor\,expansion$$
<Question.>
Is it correct to solve the Lebesgue integral like this? Or am I misunderstanding the Lebesgue integral?