Probability of uniformly distributed random variable

Question

Let X be a standard uniformly distributed random variable X.

I try to calculate $\mathbb{P}(X \in [0,\frac{1}{2}[)$

What I did:

I know that $\mathbb{P}(X \in [0,\frac{1}{2}]) = \int_0^\frac{1}{2} \frac{1}{1-0} dy = \frac{1}{2}$

$\mathbb{P}$ is a probability measure so by $\sigma$-additivity: $\mathbb{P}(X \in [0,\frac{1}{2}[) = \mathbb{P}(X \in [0,\frac{1}{2}]) -\mathbb{P}(X = \frac{1}{2}) $

But how to calculate $\mathbb{P}(X = \frac{1}{2})$ ?

Thank you for your help!

Answer 1

The standard uniform distribution is continuous, and it has no singletons with positive probability. $$\mathbb{P}\left(X = \frac{1}{2}\right) = 0$$