Is it wrong to say that a Riemann sum is an unbiased estimate of an integral? Would it be wrong to say that a Riemann sum approximation of an integral
\begin{align}
    \int_a^b f(t) \mathrm{d}t \approx \sum_{k=1}^{n_\text{samples}} f(t^{\ast}_k)\Delta t, 
\end{align}
where $\Delta t = \left(b - a\right)/n_\text{samples}$, and where $t^{\ast}_k$ is the left or right end point or the midpoint of the sub-intervals is an unbiased estimate of the true integral?
The argument for Monte Carlo integral approximation with $N$ uniform samples in the interval $[a,b]$ seems to be that as the number of samples goes to infinity, then the approximation will be the exact integral with probability 1 (see e.g https://cs.dartmouth.edu/wjarosz/publications/dissertation/appendixA.pdf, page 153)
This same limit for the Riemann sum is (the definition of) the Riemann integral, hence I would argue that a Riemann sum is also unbiased.
According to a blog post (https://blog.evjang.com/2016/09/riemann-bias.html)  which I found on Google, then the Riemann sum is biased because of the deterministic steps.
But since the argument for the Monte Carlo integration being unbiased uses that $N$ goes to infinity, I can't see why the same argument can't be used for the Riemann sum approximation.
If it indeed is wrong that the Riemann sum is unbiased, I would be happy if anyone could explain the differences in the arguments.

Any constant is a biased estimator of any different constant
Since you are using a deterministic procedure here, your Riemann sum depends only on $n$, so it is a sequence of constants.  Applying the concept of statistical bias to constants is simple --- any constant is a biased estimator of any different constant and an unbiased estimator of itself.  So, for example, $3$ is a biased estimator of $2$, but it is an unbiased estimator of $3$.
Your Riemann sum is generally going to be a biased estimator for the corresponding integral because you are selecting the points $t_k^*$ deterministically within the interval and so your estimator is a constant.  There are exceptions, for functions where the Riemann sum happens to be exactly equal to the integral (e.g., piecewise linear functions).  When the Riemann sum gives a different value to the integral it is biased (in the same way that $3$ is a biased estimator of $2$).  When the Riemann sum gives the same value as the integral it is unbiased (in the same way that $3$ is an unbiased estimator of $3$).  Irrespective of whether the Riemann sum is a biased estimator or not, it will still be a consistent estimator, since it converges to the true integral as $n \rightarrow \infty$; indeed, this is the essence of the Riemann integral.
Now, if you were to select the $t$ point uniformly at random within the interval, the resulting Riemann sum would be an unbiased estimator of the integral.  This would be a variation of estimation by importance sampling, where you are varying things by generating your points conditionally within segments of a partition.
