Let $f:[a,b] \to \mathbb{R}$ be an integrable function on $[a,b]$ and $(P_n)$ a sequence of partitions such that $\lambda(P_n)\to 0$. Is $$\lim_{n\to +\infty}U(f,P_n)=\int\limits_a^bf(x)\,\mathrm{d}x~?$$
Attempt. In case $f$ is continuous, the answer is yes. Maximums of $f$ are attained on the closed bounded intevals defined by $P_n's$, so:
$$U(f,P_n)=R(f,P_n,K_n)$$
where $R(f,P_n,K_n)$ stands for the Riemann sum defined by partition $P_n$ and choise $K_n$ of points of $P_n$, so $U(f,P_n)=R(f,P_n,K_n)\to \int\limits_a^bf$.
But what happens in the case $f$ is integrable, but not continuous?
Thanks in advance for the help.
Using the guideline by @Paramanand Singh, $n \in \mathbb{N}$. For all $k$ we can find $\xi_k^{(n)}$ such that $$\sup_{x\in [x_k^{(n)},x_{k+1}^{(n)}]}f(x)-\frac{1}{n}<f(\xi_k^{(n)})\leqslant \sup_{x\in [x_k^{(n)},x_{k+1}^{(n)}]}f(x), $$ where $[x_k^{(n)},x_{k+1}^{(n)}]$ is an interval defined by partition $P_n$. Then $$U(f,P_n)-\frac{1}{n}<R(f,P_n,K_n)\leqslant U(f,P_n),$$ or: $$ R(f,P_n,K_n)\leqslant U(f,P_n)<R(f,P_n,K_n)+\frac{1}{n}.$$ Since $R(f,P_n,K_n) \to \int\limits_a^bf$, we get $U(f,P_n) \to \int\limits_a^bf.$