Actually, this is a very general issue that affects all continuous measurable quantities. Whenever you make a measurement of a continuous quantity, you are choosing from all possible real numbers within the allowed range for that quantity. There are an infinite number of possible values, and thus the probability of getting a specific one of them - like $\xi = 0.01$, in your example - is zero. And it should make sense that the probability of getting $\xi > 0.01$ differs from the probability of getting $\xi \ge 0.01$ by the probability of getting $\xi = 0.01$: zero, which means the two former probabilities are the same.
$$\begin{align}
P(\xi > 0.01) &= P(\xi \ge 0.01) - P(\xi = 0.01) \tag{1} \\
&= P(\xi \ge 0.01) - 0
\end{align}$$
Of course, that's not a rigorous mathematical argument, just a way to intuitively understand why the two functions might be the same.
If you want a more formal demonstration, you have to consider the probability distribution function $p(\xi)$. which is defined such that the probability of $\xi$ being between $a$ and $b$ is equal to
$$\int_a^b p(\xi)\mathrm{d}\xi$$
We use this because it doesn't make sense to talk of the measured value of $\xi$ being equal to something because you can never actually establish that it is equal, only that it is close to within some precision. In your specific case, you can't talk about $\xi$ being equal to 0.01, but you can talk about it being in some small range around 0.01. So, whereas equation (1) from above doesn't really make sense, this equation does make sense:
$$\int_{0.01+\epsilon}^\infty p(\xi)\mathrm{d}\xi = \int_{0.01-\epsilon}^\infty p(\xi)\mathrm{d}\xi - \int_{0.01-\epsilon}^{0.01+\epsilon} p(\xi)\mathrm{d}\xi$$
Here $\epsilon$ is some small value which you can think of as the precision of your measuring device. If $\lvert\xi - 0.01\rvert < \epsilon$ then your measuring device will read 0.01, otherwise it will read something else. (This is a highly simplified model of measurement error, but it works for my purposes here.)
Because $p(\xi)$ is finite around 0.01, you can in principle use a measuring device that is precise enough (i.e. small enough $\epsilon$) that the last term, $\int_{0.01-\epsilon}^{0.01+\epsilon} p(\xi)\mathrm{d}\xi$, is small enough to consider negligible, no matter what your definition of "negligible" is. When you do that, then under your definition of "negligible", you get
$$\int_{0.01+\epsilon}^\infty p(\xi)\mathrm{d}\xi = \int_{0.01-\epsilon}^\infty p(\xi)\mathrm{d}\xi$$
which is exactly the statement you're looking for, that the probability of measuring $\xi > 0.01$ is the same as the probability of measuring $\xi \ge 0.01$.
