# Discrete measure and Lebesgue measurability

According to wikipedia

https://en.wikipedia.org/wiki/Discrete_measure

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 $$$$\delta_{s_i}(X)=\begin{cases} 1, & \text{if s_{i}\in X}\\ 0, & \text{if s_{i}\not\in X} \end{cases}$$$$ for any Lebesgue measurable set $$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?

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.