I am a beginner at Riemann Integration and before I delve deeper into the subject, I would like it to be verified that I understand the working processes. Here is the question:
Consider a function $f:[0,1]\to\mathbb R$ defined as follows: $f(x)=x^2$ if $x$ is rational, and $f(x)=x^3$ if $x$ is irrational. Compute $U(f),L(f)$ and check if $f$ is Riemann integrable on $[0,1]$.
My solution is as follows. Please verify if it is correct.:
First we evaluate $U(f)$ as it is straightforward. Take a partition $P=\{x_i: i=0,1,...,n\}$ with $x_i=\dfrac{i}{n}$. Now $U(P,f)=\sum \sup\{f(x):x\in[x_i,x_{i+1}]\}(x_{i+1}-x_i)=\dfrac{1}{n}\sum_0^{n-1}\dfrac{(i+1)^2}{n^2}=\dfrac{\sum_1^ni^2}{n^3}\to\dfrac{1}{3}=U(f)$
Next to evaluate $L(f)$ take a partition $P=\{\dfrac{i}{n+\sqrt{2}}:i=0,1,...,n\}\cup\{1\}$. I chose this so that $||P||\to 0$ as $P$ is refined. $L(P,f)=\sum\inf\{f(x):x\in[x_i,x_{i+1}]\}(x_{i+1}-x_i)=\dfrac{\sum_0^{n-1}\dfrac{i^3}{(n+\sqrt{2})^3}}{n+\sqrt{2}}+(\dfrac{n}{n+\sqrt{2}})^3(1-\dfrac{n}{n+\sqrt{2}})=\dfrac{(n-1)^2n^2}{4(n+\sqrt{2})^4}+(\dfrac{n}{n+\sqrt{2}})^3(1-\dfrac{n}{n+\sqrt{2}})\to\dfrac{1}{4}+0=\dfrac{1}{4}=L(f)$
Hence $U(f)\neq L(f)$ so $f$ is not Riemann Integrable on $[0,1]$.