Probability of Continuous Random Variable Lying in Singleton Set is Zero

Theorem
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability density function.

Let $X$ be a continuous real variable on $\struct {\Omega, \Sigma, \Pr}$.

Then, for each $x \in \R$, we have:


 * $\map \Pr {X \le x} = \map \Pr {X < x}$

In particular:


 * $\map \Pr {X = x} = 0$

Proof
Let $F_X$ be the cumulative distribution function of $X$ so that:


 * $\map {F_X} x = \map \Pr {X \le x}$

for each $x \in \R$.

Let $P_X$ be the probability distribution of $X$.

Since $X$ is a continuous real variable, we have:


 * $F_X$ is continuous.

From Sequential Continuity is Equivalent to Continuity in the Reals, we have:


 * for each real sequence $\sequence {x_n}_{n \mathop \in \N}$ converging to $x$, we have $\map {F_X} {x_n}\to \map {F_X} x$.

For each $n \in \N$, let:


 * $\ds x_n = x - \frac 1 n$

We have that $x_n \to x$, so:


 * $\ds \map {F_X} x = \lim_{n \mathop \to \infty} \map {F_X} {x - \frac 1 n}$

We also have that $\sequence {x_n}_{n \mathop \in \N}$ is a increasing sequence.

So, we have:


 * $\ds \hointl {-\infty} {x - \frac 1 n} \subseteq \hointl {-\infty} {x - \frac 1 {n + 1} }$

for each $n \in \N$.

So:


 * $\ds \sequence {\hointl {-\infty} {x - \frac 1 n} }_{n \mathop \in \N}$ is a increasing sequence of sets.

We can see that:


 * $\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n} = \openint {-\infty} x$

Lemma
So, we have:

We then have:

giving:


 * $\map \Pr {X = x} = 0$