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In this question Why does the monotone convergence theorem not apply on Riemann integrals? there is the typical counterexample showing that the monotone convergence theorem does not hold for Riemann integrals. My question is: Why is $f_n(x)$ Riemann integrable for all $n$?

From the Lebesgue criterion, a function is Riemann integrable if it is bounded and the set of discontinuites has Lebesgue measure $0$. The boundedness is clear, however, I thought that $f_n(x)$ would be discontinuous for all $x \in [0,r_{n-1}]$ (assuming the enumeration of the rationals $r_1,r_2,...$ is ordered from lowest to highest) since if $x$ is an irrational number in $[0,r_{n-1}]$ I can find a sequence $x_k$, $k = 1,2,...$ in $[0,r_{n-1}]$ of rational numbers such that $f_n(x_k) = 1$ for all $k$ and $f_n(x) = 0$. And the reverse argument if $x \in [0,r_{n-1}]$ is rational. Finally the Lebesgue measure of $[0,r_{n-1}]$ is $r_{n-1}$ which is not $0$.

Also the upper Riemann sums should be strictly greater than zero since for some interval $[x_i, x_{i+1}]$ it will happen that $\sup_{x \in [x_i, x_{i+1}]} f_n(x) = 1$ (since rationals are dense in the reals) while the lower Riemann sum will always be zero.

What am I missing in these arguments?

improbable_probabilist
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    If you modify a Riemann-integrable function in one point (and therefore in finitely many points) - it remains Riemann-integrable. So, we have obtained $f_n(x)$ by taking the zero function and just modifying it at $r_0, r_1, \ldots , r_{n-1}$ to become $1$ (instead of $0$) - it stays Riemann-integrable. –  Aug 17 '22 at 11:51

2 Answers2

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For each fixed $n$ the function $f_n$ is $0$ except at a finite number of points $r_0,r_1,..,r_{n-1}$. Such a function is clearly continuous at all points except this finite number, i.e. $r_0,r_1,..,r_{n-1}$. [You get $n$ open intervals when you remove these points from the domain and $f_n$ is continuous on each of these open intervals].

David C. Ullrich
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Kavi Rama Murthy
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  • but since rationals are dense in the reals any open interval will contain rational numbers, hence there you cannot take $n$ open intervals without any rational numbers, no? – improbable_probabilist Aug 17 '22 at 12:00
  • I guess what i am not seeing is that in $[0,r_{n-1}]$ the set of rationals is countable (infintely), however the counterexample refers to a finite enumeration and hence enumerates just some rationals leaving space for open intervals without any rationals – improbable_probabilist Aug 17 '22 at 12:04
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    You are forgetting that $f_n$ does not involve all rationals number at all. Its definition involves only $n$ rational numbers. Remember that you are fixing $n$. @JRT – Kavi Rama Murthy Aug 17 '22 at 12:08
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The existing comments and answers tend to explain why the result is true, with less that I can see on where you error is.

It's not true that $f_n$ is discontinuous at every rational in $[0,r_n]$. In fact $f_n$ is continuous everywhere except at $r_1,\dots,r_n$. Your $f_n(x_k)=1$ only holds when $k\le n$, so you can't let $k\to\infty$ (with $n$ fixed) the way you do.

David C. Ullrich
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