Comparison between lebesgue integral and riemann integral of $f(x)=x^2$ in $[0,2]$ If we have an example $f(x)=x^2$ let's say for $[0,2]$. 
In lebesgue integral, I already use a sequence of function $f_n(x)$ as approximation to $f(x)$ ($f_n(x)$ converges to $f(x)$) which is stated by $f_n(x)=\sum_{k=1}^{2.2^n}\frac{k-1}{2^n}1_{\left\{\left(\sqrt{\frac{k-1}{2^n}},\sqrt{\frac{k}{2^n}}\right]\right\}}$. Then we know that the step after that is by using monotone convergence theorem which say $\lim_{n\rightarrow\infty}\int_{R}f_n=\int_{R}f$. Then I got the limit of integral of $f_n$ as $\lim_{n\rightarrow\infty}\sum_{k=1}^{2.2^n}\frac{k-1}{2^n}(\sqrt{\frac{k}{2^n}}-\sqrt{\frac{k-1}{2^n}})$. 
When I try to figure out this limit of sequence of function. It just ended up not in the right answer it should be (the answer in Riemann integral is : $\frac{8}{3}$. So by using lebesgue integral, we must have the same answer as riemann integral is). 
Please help for proof.
 A: Since $f(2) = 2^2 = 4$, the upper limit for the sum should be $4 \cdot 2^n$:
$$f_n(x) = \sum_{k=1}^{4 \cdot 2^n}\frac{k-1}{2^n}1_{E_n},$$
where
$$E_n = \left\{x : \frac{k-1}{2^n} \leqslant f(x)\leqslant \frac{k}{2^n}\right\}=\left\{x : \sqrt{\frac{k-1}{2^n}} \leqslant x\leqslant \sqrt{\frac{k}{2^n}}\right\}.$$
Then the Lebesgue integral is
$$\int f= \lim_{ n \rightarrow \infty}\sum_{k=1}^{4 \cdot 2^n}\frac{k-1}{2^n}m(1_{E_n})$$
with
$$\sum_{k=1}^{4 \cdot 2^n}\frac{k-1}{2^n}m(1_{E_n})=\sum_{k=1}^{4 \cdot 2^n}\frac{k-1}{2^n}\left(\sqrt{\frac{k}{2^n}}-\sqrt{\frac{k-1}{2^n}}\right)\\=\frac{1}{2^n\sqrt{2^n}}\sum_{k=1}^{4 \cdot 2^n}(k-1)\left(\sqrt{k}-\sqrt{k-1}\right).$$
We can simplify this using summation by parts:
$$\sum_{k=1}^{m}a_k(b_{k+1}-b_k)=  a_{m+1}b_{m+1}-a_1b_1 - \sum_{k=1}^{m}b_{k+1}(a_{k+1}-a_k)$$
Summing by parts, we obtain
$$\frac{1}{2^n\sqrt{2^n}}\sum_{k=1}^{4 \cdot 2^n}(k-1)\left(\sqrt{k}-\sqrt{k-1}\right) \\=\frac{1}{2^n\sqrt{2^n}}\left[4 \cdot 2^n\sqrt{4 \cdot 2^n}-\sum_{k=1}^{4 \cdot 2^n}\sqrt{k}\right]\\=8-\frac{1}{2^n}\sum_{k=1}^{4 \cdot 2^n}\sqrt{\frac{k}{2^n}}.$$
Hence,
$$\int f= 8-\lim_{ n \rightarrow \infty}\frac{1}{2^n}\sum_{k=1}^{4 \cdot 2^n}\sqrt{\frac{k}{2^n}}.$$
The limit on the RHS can be recognized as the limit of a Riemann sum for $\sqrt{x}$ on the interval $[0,4]$.
Whence,
$$\int f= 8-\int_0^4\sqrt{x} \, dx = 8-\left.\frac{2}{3} x^{3/2}\right|_0^4=\frac{8}{3}$$
