Suppose ${P_n}$ is a sequence of partitions of $[a,b]$, each of which is refined by its successor (i.e., $\forall n \in \mathbb N, P_n \subseteq P_{n+1} $). Show that for any bounded $f :[a,b]\to \mathbb R$, integrability of $f$ on $[a,b]$ does not guarantee that $\lim\limits_{n \to \infty}\underline{S}(f,P_n)=\lim\limits_{n \to \infty}\overline{S}(f,P_n)$.
We can show it by counterexample. Take function $f(x)=x$ on $[0,2]$ and $P_n=\{0,\frac{1}{2^n}, \frac{2}{2^n},....,\frac{2^n}{2^n},2\}$. From this I get $\lim\limits_{n \to \infty}\underline{S}(f,P_n) = 3/2$ and $\lim\limits_{n \to \infty}\overline{S}(f,P_n)= 5/2$. These two are not equal.
As $f(x)=x$ is a continuous function on $[0,2]$, so is integrable on $[0,2]$. I have studied that if $f$ is integrable on $[a,b]$ and there exist a sequence $\{P_n\}$ of partitions of $[a,b]$ such that $\underline{S}(f,P_n)\to L$ and $\overline{S}(f,P_n)\to M$, then $L=M=\int_{a}^bf$. But why here we get a contradiction in the solution to the above question as we get a sequence of partition such that $\underline{S}(f,P_n)$ and $\overline{S}(f,P_n)$ converges but to different numbers? Why this happens?
By Cauchy criterion, $f$ is integrable at $[a,b]$ if and only if there exists a sequence $ (P_n) $ of partitions of $[a,b]$ such that $$\lim_{n\to+\infty}(U(f,P_n)-L(f,P_n))=0$$
The sequence you chosen does not necessarily satisfy that.