Lebesgue integral of ${x^2}$

Question

I am having trouble calculating the Lebesgue integral of ${x^2}$ over ${]0,1]}$. I followed the instructions from my college lessons and constructed a sequence ${f_n}:=\sum_{k=1}^{n*2^n}\frac{k-1}{2^n}*\chi{}_{A_{n,k}}$ with ${A_{n,k}:=\{ \sqrt{\frac{k-1}{2^n}} \leq x < \sqrt{\frac{k}{2^n}} \}}$. ${\lambda(f)}$ is equal to ${sup_{n\in N}\lambda(f_n)}$, so I calculated $\lambda(f_n)=\sum_{k=1}^{n*2^n}\frac{k-1}{2^n} * (\sqrt{\frac{k}{2^n}} - \sqrt{\frac{k-1}{2^n}})$. However, I don't see how this is going to be equal to ${\frac{1}{3}}$, which would be the result of a Riemann integral. Can anyone spot a mistake or point me in the right direction?

Answer 1

Try using $A_{n,k}:=\left\{ \frac{k-1}{2^n} \leq x < \frac{k}{2^n} \right\}$ and

$$f_n = \sum_{k=1}^{2^n} \left(\frac{k-1}{2^n}\right)^2 \chi_{ A_{n,k}}$$

Since $f_n \leqslant f_{n+1}$ and $f_n(x) \to f(x) =x^2$ pointwise for $x \in [0,1]$, we have by the monotone convergence theorem,

$$\begin{align}\int_0^1 x^2 \, dx = \int_{[0,1]} f &= \lim_{n \to \infty} \int_{[0,1]} f_n \\ &= \lim_{n \to \infty} \sum_{k=1}^{2^n} \frac{(k-1)^2}{2^{2n}} \left(\frac{k}{2^n} - \frac{k-1}{2^n} \right) \\ &= \lim_{n \to \infty} \frac{1}{2^{3n}}\sum_{k=0}^{2^n-1} k^2 \\ &= \lim_{n \to \infty} \frac{2^n(2^{n}-1)(2\cdot 2^n -1)}{2^{3n}\cdot 6} \\ &= \frac{1}{3}\end{align}$$