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.


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.


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    $\begingroup$ Can you use the Lebesgue theorem that states that monotone functions on closed, bounded interval are differentiable almost everywhere, and hence continuous almost everywhere, which proves Riemann integrability? $\endgroup$ – Sawyer R. Jan 2 '18 at 20:08
  • $\begingroup$ That would have worked yes. $\endgroup$ – RhythmInk Jan 2 '18 at 20:10
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    $\begingroup$ The problem is odd because the hypotheses are unnecessary. Any monotone function is Riemann integrable. $\endgroup$ – AlohaSine Jan 2 '18 at 20:12
  • $\begingroup$ Agreed with mathstudent, the sequence is irrelevant. $\endgroup$ – Sawyer R. Jan 2 '18 at 20:13

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.

  • $\begingroup$ Why is that set countable? $\endgroup$ – RhythmInk Jan 2 '18 at 20:10
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    $\begingroup$ @az89 Because a monotone function can only have jump discontinuities. And it is easy to see that (since it is bounded), for each natural $n$ there can be only be finitely many points $a\in[0,1]$ such that the jump (that is $\lim_{x\to a^+}f(x)-\lim_{x\to a^-}f(x)$) is geater than or equal to $\frac1n$. $\endgroup$ – José Carlos Santos Jan 2 '18 at 20:13
  • $\begingroup$ @SawyerR. Are you sure that it is false for monotone functions? $\endgroup$ – José Carlos Santos Jan 2 '18 at 20:13
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    $\begingroup$ The set of discontinuities of a monotone function from an interval to $\mathbb{R}$ is definitely countable. $\endgroup$ – Ian Jan 2 '18 at 20:14
  • $\begingroup$ I was definitely wrong, I recall the proof now. I was getting that fact confused with riemann integrable iff set of discontinuities has measure zero. $\endgroup$ – Sawyer R. Jan 2 '18 at 20:15

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

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


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

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


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