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I know a proof that Riemann integrable positive function $f$ defined on $[a,b]$ is bounded.

Suppose $f\geq 0$ is unbounded. First, for each $n$, choose a point $a_n$ in $[a,b]$ such that $f(a_n) > n$. Let $a$ be an arbitrary positive real number. Choose $n$ with $a > \frac1n$, and we can have a partition $p$ with each interval length bigger than $\frac1n$ while mesh($p$) $< a$. Then $R(f, p) > f(a_{n^2}) \frac1n = n$ so the Riemann sums do not tend to a finite limit.

But I think while we choose $a_n$ in the first step, countable choice is used. Are there any other ways to prove it without using some kind of choice, since this is a very 'basic' theorem in 'elementary 'analysis so it seems weird that it requires some kind of choice axiom.

Arctic Char
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Sphere
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    Riemann integrable functions on $[a,b]$ are bounded by definition. If you are considering improper Riemann intergals then you cannot prove that $f$ is bounded. – Kavi Rama Murthy Nov 05 '20 at 07:46
  • A Riemann-integrable function is a uniform limit of steps functions. Step functions are bounded, so the (uniform) limit is also. So Riemann integrable functions are bounded. – TheSilverDoe Nov 05 '20 at 07:48
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    I am studying real analysis and foundations by Krantz. And it defines the riemann integral as the limit of riemann sums as mesh tends to $0$, riemann integrable funciton as function which the limit exists. – Sphere Nov 05 '20 at 07:55
  • Another way is to argue that, for any partition, the supremum of $f$ on some some subinterval must be infinite and so will be the upper Darboux sum. Otherwise we can show that no real number can be the limit of Riemann sums as proved here more generally for the Riemann-Stieltjes integral. This should avoid the problem you describe. – RRL Nov 05 '20 at 08:21

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You don't have to choose the whole sequence in advance. We assumed that $f$ is unbounded, so for any given $n$ we can find $x$ such that $f(x) > n^2$, that's just part of the definition.

Now for any fixed $a$ pick a suitable $n$, and $x$ such that $f(x)>n^2$, etc. We are really using the whole sequence here. It's a proof of convenience.

Asaf Karagila
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Suppose $f:[a,b]\to\mathbb{R}$ is Riemann integrable but assume $f$ is unbounded. Let $I_R=\int_{a}^{b}f$, which is also defined as $\lim_{||P||\to 0}R(f,P,x^*)$, where $P=\{a=t_0<t_1<...<t_n=b\}$ is a partition of $[a,b]$ and $x^*$ is a compatible n-tuple with $P$, that is, $x_i^*\in[t_{i-1},t_i]$ for all $1\leq i\leq n$. Also, $||P||$ is the norm of $P$, which is the maximum of $t_i-t_{i-1}$.

Since $f$ is integrable, then for all $\varepsilon>0$, there exists a $\delta>0$ such that for all partitions $P$ with $||P||<\delta$, then $|R(f,P,x^*)-I_R|<\varepsilon$.

Consider the particular choice $\varepsilon=57$. Then there exists a $\delta>0$ such that for all partitions $P$ with $||P||<\delta$, we have $|R(f,P,x^*)-I_R|<57$.

If $x^*$ and $z^*$ are compatible with $P$, then we have $$\begin{align*}|R(f,P,x^*)-R(f,P,z^*)|&= |R(f,P,x^*)-I_R+I_R-R(f,P,z^*)|\\&\leq|R(f,P,x^*)-I_R|+|I_R-R(f,P,z^*)|\\&<114\end{align*}$$

This is a set up for the following, where we will redefine $x^*$ and $z^*$.

By the Archimedean property, choose $n\in\mathbb{N}$ such that $P_n$ is a regular partition of $[a,b]$ so that if $P_n=\{a=t_0<t_1<...<t_n=b\}$, then $t_i-t_{i-1}=\frac{b-a}{n}$ and $||P_n||<\delta$.

Suppose $x^*$ is compatible with $P_n$. Since $f$ is unbounded, then there exists some $j\in\mathbb{N}$ with $1\leq j\leq n$ such that $f|_{[t_{j-1},t_j]}:[t_{j-1},t_j]\to\mathbb{R}$ is unbounded.

In particular, there exists a $z_j\in[t_{j-1},t_j]$ such that $|f(z_j)-f(x_j^*)|>\frac{570n}{b-a}$.

Now, define $z^*$ as $x_i^*$ if $i\neq j$ and $z_j$ if $i=j$. Then we have $$\begin{align*}|R(f,P_n,x*)-R(f,P_n,z^*)|&=\big|\sum_{i=1}^{n}f(x_i^*)(t_i-t_{i-1})-\sum_{i=1}^n f(z_i^*)(t_i-t_{i-1})\big|\\&=|[f(x_j^*)-f(z_j)](t_j-t_{j-1})|\\&>\frac{570n}{b-a}\cdot\frac{b-a}{n}\\&=570 \end{align*} $$

However, we have shown that this quantity should be less than $114$, therefore we have reached a contradiction. Thus $f$ must be bounded.

C Squared
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