Lebesgue measurable function Riemann integrable

Question

Problem

Let $f_n:[0,1] \to \mathbb{R}$ be Lebesgue measurable functions. Suppose they converge point-wise to a monotone function $f:[0,1] \to \mathbb{R}$. Prove that $f$ is Riemann integrable.

Attempt

My first thought is to use Lebesgue-vitali theorem, but I don't have anything about boundedness or continuity. I'm wondering if maybe I need to do it directly showing the upper and lower sum difference converges.

Thanks!

Answer 1

Since $f$ is monotonic and the domain is a closed and bounded interval, $f$ must be bounded. And the set of points at which it is discontinuous is countable, and therefore its Lebesgue measure is $0$. So, $f$ is Riemann-integrable.

Answer 2

Here's an elementary proof that every monotone function on $[0,1]$ is Riemann integrable. Suppose $f$ is nondecreasing on $[0,1].$ Consider the uniform partition $P_n$ of $[0,1]$ with $n$ subintervals. Then

$$U(f,P_n) = \sum_{k=1}^{n}f\left(\frac{k}{n}\right)\frac{1}{n},\,\,\, L(f,P_n) = \sum_{k=1}^{n}f\left(\frac{(k-1)}{n}\right)\frac{1}{n}.$$

Thus

$$U(f,P_n)-L(f,P_n) = f(1)\frac{1}{n}-f(0)\frac{1}{n}=\frac{f(1)-f(0)}{n}.$$

Therefore $U(f,P_n)-L(f,P_n)\to 0$ as $n\to \infty,$ hence $f$ is Riemann integrable on $[0,1].$