Constructing a Measure on the Rational Numbers

No, your suggestion is not a probability measure as it is not additive over countable unions of disjoint sets.

If $t$ is a natural number then $m(\{q_t\})= \lim_{n \to \infty} \frac1n=0$ and so $\sum_t m(\{q_t\})=0$. But $m( \mathbb{Q})= \lim_{n \to \infty}\frac{n}{n}=1$.

You need a discrete measure.

For example $\mathbb{Q} \cap (0,1)$ you could use $m \left( \left\{ \frac{a}{b} \right\}\right) =\frac{\zeta(k)}{\zeta(k-1) - \zeta(k) } \left(\frac{1}{b}\right)^k$ for some $k\gt2$ and $a$ and $b$ coprime; this can be extended to all rationals.

A simple example of such a measure is the following, but this is a boring example.

Let $\{ q_n \}_{n=1}^ \infty$ be an enumeration of the rationals.

Define $$m=\sum_{k=1}^\infty \frac{1}{2^k} \delta_{q_k} \,.$$ This measure can be defined alternatelly $$m(E) = \sum_{ q_n \in E} \frac{1}{2^n}$$

Note that $m({q}) = 0$ for every $q \in \mathbb{Q}$, whereas $m(\mathbb{Q}) = 1$, so $m$ cannot be $\sigma$-additive. This argument shows that there are no probability measures on a denumerable set that assign the same measure to every singleton.

Moreover, the limit in your definition does not always exist.

There are, however, interesting examples of finitely additive measures (or probability charges). The example you mention if in fact to being a probability charge, but one must be careful enough to consider the cases where the limit does not exist see this article.