Proof that the Riemann-sums of $\int_0^1\tfrac{1}{x}\text{d}x$ diverge to $+\infty$

Question

Define the function $f:[0,1]\to\mathbb R$ by $$f(x) = \left\{\begin{matrix} 0 \qquad &\text{if} &\qquad x=0\\ 1/x \qquad &\text{if} &\qquad x\neq 0. \end{matrix}\right.$$

$f$ is not bounded, so it is not Riemann integrable. However, we can still try to make sense of the limit of the Riemann sums. It seems very likely to me that the Riemann sums will converge to $\infty$, with the limit taken in the same sense as for the Riemann integral of Riemann-integrable functions, namely, that for every sequence of tagged partitions ${D_n}$ of $[0,1]$ such that the mesh size $\nu(D_n)\to 0$ as $n\to\infty$ we have that the limit of the Riemann-sums satisfies $$\lim_{n\to\infty}\sum_{i=1}^{m}f(\xi_i)\Delta x_i = \infty.$$ Here I have represented the tagged partition $D_n$ as $D_n = \{x_0=0\leq \xi_1\leq x_1\leq\dots\leq \xi_m\leq x_m = 1 \}$ (where all $\xi_i$ and $x_i$, and also the $m$ appearing in the sum, depend on $n$, but I have ommitted this for clarity).

Question: Is this correct? And most importantly, how do we prove this?

I'm not 100% sure the result is true but it seems very likely to me. In that case it would make sense to say that $\int_0^1f(x)\text{d} x = \infty$ even though the function is not Riemann-integrable. I have been trying to prove it but did not succeed yet. It's tempting to just use the fact that $\int_{1/m}^1f(x) = -\log(1/m)$ and then take the limit $m\to\infty$. However, this is a different definition of the integral, and it is not obvious at all that it should be equivalent to the definition of the integral that I gave above.

My (unsuccessful) idea was to split the integral as $\int_0^1f(x)\text{d} x = \int_0^{\lambda}f(x)\text{d} x + \int_\lambda^1f(x) \text d x$ (this is symbolic, we should do this split in the Riemann sums), taking a small $\lambda$. Then if we have a partition of $[0,1]$ containing $\lambda$ as one of its points $x_k$, and with sufficiently small mesh size, then the second integral (actually the corresponding Riemann sum) becomes arbitrarily close to $-\log\lambda$, and the first integral (actually the corresponding Riemann sum) is always positive, so the complete expression is always larger than $-\log\lambda$. Hence taking $\lambda\to 0$ this would seem to indicate that we can make the Riemann sums arbitrarily large, hence the integral would converge to $\infty$. The problem with this argument is of course that I have assumed beforehand that the partition contains $\lambda$ as one of its points $x_k$. We need the limit to give the same result for any partition, however, and not only for those ones containing $\lambda$. Now, it is true that $\lambda$ was arbitrary, so our conclusion that the total Riemann sum can be made arbitrarily large holds in some sense for all partitions, but the $\delta$ that we need in order to achieve this depends on $\lambda$, which is not what we want.

Any ideas?

Answer 1

I think you were close to a solution with the idea of splitting the domain. To fill the gap: suppose you have some $R > 0$. Then there exists $\delta_0 > 0$ such that $-\log(\delta_0) \ge 2 R$. Now, for any partition $(x_i, x_i^*)$ with norm less than $\delta_0$, we can form an "intersection partition" where the first interval starts at $\delta_0$, we take the subset of interval endpoints which are greater than $\delta_0$, and we choose either the corresponding $x_i^*$ from the original partition or replace -- or if that corresponding $x_i^*$ for the original interval containing $\delta_0$ was less than $\delta_0$, then we replace that one with $\delta_0$. The Riemann sum for the original partition is greater than the Riemann sum for the intersection partition of $\int_{\delta_0}^1 \frac{dx}{x} = -\log(\delta_0)$; and the norm of the intersection partition is at most the norm of the original partition. Since $\frac{1}{x}$ is continuous on $[\delta_0, 1]$, the limit of these lower bounds for the full Riemann sum, as $\delta \to 0^+$, is equal to $-\log(\delta_0) \ge 2 R$. Therefore, for small enough $\delta > 0$, we can conclude that the full Riemann sum is greater than $R$ for any partition of $[0, 1]$ with norm less than $\delta$.

(Note that this argument made essential use of the fact that $f$ is positive and decreasing on $(0, 1]$. It could be interesting to see whether there might be a situation for something like $\int_0^1 \frac{1}{x^2} \sin(1/x)\,dx$ where the integral diverges, but where some subset of Riemann sums might converge. Maybe by having the lower part of the partition always having $x_i^*$ being of the form $\frac{1}{n\pi}$ with $n\in \mathbb{N}$, and taking splitting points also at $\frac{1}{m\pi}$ so that the integral from that point to 1 is equal to 0?)

Answer 2

This answer is essentially just Daniel's answer (although with some small non-essential changes), but formulated in a way that I (the poster of the question) find easier to understand. I typed it out for myself anyway, so I thought why not post it here, in case it might be useful to anybody.

Let $M>0$. We will prove that there exists $\delta>0$ such that for all partitions $D$ with mesh size $\nu(D)<\delta$ the Riemann sum satisfies $\sum_{i=1}^m f(\xi_i)\Delta x_i > M$.

First, pick $\delta_0>0$ such that $-\log\delta_0\geq 2M$ (this will not be the final $\delta$). Consider any tagged partition $D$ of $[0,1]$ with mesh size $\nu(D)<\delta_0$. Recall that $\nu(D) = \max_i \Delta x_i$, by definition (I use the notation introduced in the question). Now, starting from $D$ (made up out of $x_i$ and $\xi_i$), we may define a partition $\tilde D$ of $[\delta_0,1]$ by adding the point $\delta_0$ to the $x_i$'s (if one of the $x_i$'s is already equal to $\delta_0$ this is not necessary) and removing all points $x_i$ that are smaller than $\delta_0$.

Say $\delta_0\in(x_{k-1},x_k)$. We leave the tags $\xi_i$ in each sub-interval as they are, except the tag $\xi_k\in(x_{k-1},x_k)$, which we replace by $x_k\geq \xi_k$. Then the original Riemann sum over $D$ satisfies \begin{align} \sum_{i=1}^m f(\xi_i)\Delta x_i &= \sum_{i=1}^{k-1} f(\xi_i)\Delta x_i + f(\xi_{k})\Delta x_{k} + \sum_{i=k+1}^m f(\xi_i)\Delta x_i \tag{1}\\ % & \geq f(\xi_{k})\Delta x_{k} + \sum_{i=k+1}^m f(\xi_i)\Delta x_i \tag{2}\\ % &\geq f(x_k)(x_k-\delta_0) + \sum_{i=k+2}^m f(\xi_i)\Delta x_i \tag{3} \end{align} where we have used that all terms in the sum are non-negative, and that $\xi_k\leq x_k$, $\Delta x_k\geq x_k-\delta_0$ and $f$ is decreasing on $(0,1]$, so that $f(\xi_k)\Delta x_k\geq f(x_k)(x_k-\delta_0)$. For this argument it is important that $\xi_k\neq 0$. This follows from the fact that $\Delta x_1\leq \nu(D)<\delta_0$, by assumption, so that the point $\delta_0$ can never be contained in the first sub-interval of the original partition. Since $\xi_k$ lies in the same sub-interval as $\delta_0$, we must have $\xi_k>0$.

The expression (3) is precisely the Riemann-sum of the integral $\int_{\delta_0}^1f(x)\text{d} x$ with respect to the partition $\tilde D$. This integral exists because $x\mapsto 1/x$ is continuous and bounded on $[\delta_0,1]$. By choosing the mesh size $\nu(D)$ sufficiently small, we can make $\nu(\tilde D)$ arbitrarily small, because $\nu(\tilde D)\leq \nu(D)$ by construction, and consequently we can make the expression (3) arbitrarily close to $\int_{\delta_0}^1f(x)\text{d} x= -\log(\delta_0)\geq 2M$. More precisely, given $\varepsilon>0$ there exists $\delta>0$ (and we may assume wlog that $\delta<\delta_0$) such that if $\nu(D)<\delta$ then \begin{align} \left|f(x_k)(x_k-\delta_0) + \sum_{i=k+2}^m f(\xi_i)\Delta x_i + \log(\delta_0)\right|<\varepsilon, \end{align} and hence in particular, \begin{align} f(x_k)(x_k-\delta_0) + \sum_{i=k+2}^m f(\xi_i)\Delta x_i &> -\log(\delta_0) - \varepsilon \\ &\geq 2M - \varepsilon \\ &> M, \end{align} where in the last step we have assumed wlog that $\varepsilon<M$. By Eq. (1-3) it therefore follows that \begin{align} \sum_{i=1}^m f(\xi_i)\Delta x_i > M. \end{align} This holds for all partitions $D$ of $[0,1]$ with $\nu(D)<\delta$. Since $M$ was arbitrary, we can make the Riemann sum arbitrarily large by choosing the partitions sufficiently fine. Hence, by defintion, as $\nu(D)\to 0$ the Riemann sums tend to $\infty$, as desired.