Define, with $x$ real valued:
$$
F_N(x) = \prod_{k=0}^N\left|x-\frac{k}{N}\right|^{1/N}
$$
We can calculate the limit for $N\to\infty$ as a Riemann sum converging to an integral:
$$
F_\infty(x) = \lim_{N\to\infty} \prod_{k=0}^N\left|x-\frac{k}{N}\right|^{1/N} = \\
\lim_{N\to\infty} \exp\left[\frac{1}{N}\sum_{k=0}^N\ln\left|x-\frac{k}{N}\right|\right] = \\
\exp\left[\int_0^1\ln|x-t|dt\right] \quad \Longrightarrow \\
F_\infty(x) = \exp\left[(x-1)\ln|x-1| + x\ln|x|-1\right]
$$
It can easily be shown that $\,F_\infty(0)=F_\infty(1)=1/e\,$ and there is a minimum for $\,F_\infty(1/2)=1/e/2\,$.
A graph says more than a thousand words:
The interval $\,[0,1]\,$ and the minimum of $\,F_\infty\,$ are delimited with red lines.
A sample function $\,F_N(x)\,$ for $\,N=9\,$ is displayed in black. The limit function $\,F_\infty(x)\,$ is displayed in green.
Numerically it is suggested that the black curves $\,F_N\,$ converge to the green curve $\,F_\infty\,$.
With exception, though, of the function values $\,f_N(k/N)\,$; all of these must be zero from the start.
But in the limit this would mean that $\,F_\infty(x)\,$ is zero as well as non-zero at the whole $\,[0,1]\,$ interval.
Which is of course impossible.
What is a sensible way out of this dilemma?
The answer is simple: $F_N(x)$ does not converge to $\exp\left(\int_{0}^{1}\log|x-t|\,\mathrm{d}t\right)$ when $x \in \mathbb{Q}\cap[0,1]$.
Let $x \in [0, 1]$. Then it is not hard to check that
$$ \log F_N(x) = \int_{0}^{1} \log|x-t| \, \mathrm{d}t + \frac{1}{N} \log\operatorname{dist}\biggl( x, \frac{1}{N}\{0,1,\ldots,N\} \biggr) + \mathcal{O}\biggl( \frac{\log N}{N} \biggr). $$
So, $\log F_N(x)$ converges to $\int_{0}^{1} \log|x-t| \, \mathrm{d}t$ if and only if
$$ \liminf_{N\to\infty} \frac{1}{N} \log\operatorname{dist}\biggl( x, \frac{1}{N}\{0,1,\ldots,N\} \biggr) = 0. $$
Equivalently, $\log F_N(x)$ does not converge to $\int_{0}^{1} \log|x-t| \, \mathrm{d}t$ if and only if
$$ x \in A := \bigcap_{p=1}^{\infty}\bigcap_{N=1}^{\infty}\bigcup_{n=N}^{\infty}\bigcup_{k=0}^{n} \biggl( \frac{k}{n}-e^{-n/p}, \frac{k}{n}+e^{-n/p}\biggr). $$
It is also interesting to note that $A$ is a $G_{\delta}$-set containing $\mathbb{Q}\cap[0,1]$, hence is uncountable.