If $f$ is Riemann integrable on a closed interval, does its integral exists for any sequence of partitions with norm converges to $0$?

Question

It can ben shown that if $f$ is continuous on $[a, b]$ then $\int_a^bf\ dx=\lim_{n\to\infty}\ U(f, P_n)=\lim_{n\to\infty}\ L(f, P_n)$ for any sequence of partitions of $[a, b]$ with norm (or mesh) converges to $0$.

Does the statement still hold if the continuity condition of $f$ is replaced by $f$ is Riemann integrable?

I find the statement still holds for $f$ being discontinuous at finitely many points so if it doesn't hold then any function for a counter-example has to be discontinuous at infinitely many points.

Answer 1

Based on the comments and using the following definition of Riemann integral:

We say that a bounded function $f: [a, b]\to\mathbb{R}$ is integrable on $[a, b]$ if $U(f) = L(f)$. In this case we use the symbols $$ \int_a^b\ f(x)\ dx\hspace{1em}\text{or}\hspace{1em}\int_a^b\ f\ dx $$ to represent this common value.

I came up with the following sketch of proof.

Let $f$ be integrable on $[a, b]$ and $(P_n)$ be a sequence of partitions of $[a, b]$ with norm converges to $0$. We'll need to show that the following string of equalities hold:

$$ \int_a^b\ f\ dx=\lim_{n\to\infty}\ U(f, P_n)=\lim_{n\to\infty}\ L(f, P_n). $$

From the fact that $f$ is integrable and the definitions of supremum and infimum, it can be shown that there exists a sequence of partitions $(Q_n)$ such that:

$$ \int_a^b\ f\ dx=\lim_{n\to\infty}\ U(f, Q_n)=\lim_{n\to\infty}\ L(f, Q_n), $$

as $\lim_{n\to\infty}\ ||Q_n||\to 0$.

Let $Q_n=\{x_0, x_1,..., x_{n_Q}\}$ where $x_{i-1}\lt x_i$ for $i=1, 2,..., n_Q$ and $M_k=\inf\{f(x): x\in[x_{k-1}, x_k]\}$, $m_k=\sup\{f(x): x\in[x_{k-1}, x_k]\}$. Then it can be shown that for each $M_k$, there exists $x^{(M)}_k\in [x_{k-1}, x_k]$ such that $f(x^{(M)}_k)=M_k$ or $\lim_{x\to x^{(M)}_k}f(x)=M_k$. Similarly, for each $m_k$, there exists $x^{(m)}_k\in[x_{k-1}, x_i]$ such that $f(x^{(m)}_k)=m_k$ or $\lim_{x\to x^{(m)}_k}f(x)=m_k$. Now for each $Q_n$, let

$$ d=\underset{1\le k \le n_Q}\min\{x^{(M)}_k-x_{k-1}, x_k-x^{(M)}_k, x^{(m)}_k-x_{k-1}, x_{k-1}-x^{(m)}_k\}. $$

Next, choose $N\in\mathbb{N}$ such that $||P_m||\lt d/2$ for all $m\gt N$. This number $N$ exists since $\lim_{n\to\infty}\ ||P_n||=0$. With such choice of $m$, it can be shown that there exist $y_{n_i}, y_{n_j}\in P_m$ such that $x_{k-1}\lt y_{n_i}\lt x^{(M)}_k\lt y_{n_i}\lt x_k$. And similarly, there exist $y'_{n_i}, y'_{n_j}\in P_m$ such that $x_{k-1}\lt y'_{n_i}\lt x^{(m)}_k\lt y'_{n_i}\lt x_k$.

From here, it's easy to see that $L(f, Q_n)\le U(f, P_m)\le U(f, P_m)\le U(f, Q_n)$. As $n\to\infty$, by Squeeze Theorem, we obtain

$$ \lim_{m\to\infty}\ U(f, P_m)=\lim_{m\to\infty}\ L(f, P_m)=\int_a^b\ f\ dx, $$

completing our proof.