According to wikipedia
a driscrite measure is defined in the following way:
Let's consider a real line $\mathbb{R}$. For some (possibly finite) sequences $s_{1}, s_{2}, \dots$ and $a_{1}, a_{2}, \dots$, s.t. $a_{i}>0$ and $\sum_{i}a_{i} = 1$, let$$ \begin{equation} \delta_{s_i}(X)=\begin{cases} 1, & \text{if $s_{i}\in X$}\\ 0, & \text{if $s_{i}\not\in X$} \end{cases} \end{equation} $$for any Lebesgue measurable set $X$. Then$$ \mu = \sum_{i}a_{i}\delta_{s_i} $$is the discrete measure on $\mathbb{R}$.
The question: why do we need Lebesgue measurability?
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$\begingroup$You don't. Such a measure makes sense on any $\sigma$-algebra.
The authors of that page probably have in mind a context where one is studying some larger class of measures on the Lebesgue $\sigma$-algebra (or perhaps the Borel $\sigma$-algebra), and where one wants to say something about which measures in that class are discrete.
Note that they discuss a more general setting further down - they want to call a measure $\mu$ discrete with respect to $\nu$ only when it is of the form you describe, and all the singletons $\{s_i\}$ are $\nu$-null. So this part requires saying something about the $\sigma$-algebra $\Sigma$ on which $\nu$ is defined. The measure $\mu$ will still make sense and behave well for sets $X$ which are not $\Sigma$-measurable, but in this context we are just not interested in such sets.
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