# Cumulative Distribution Function is Right-Continuous

## Theorem

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

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

Let $F_X$ be the cumulative distribution function of $X$.

Then:

$F_X$ is right-continuous.

## Proof

Let $x \in \R$.

We show that $F_X$ is right-continuous at $x$.

We use Sequential Right-Continuity is Equivalent to Right-Continuity in the Reals: Corollary, and will show that:

for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, that converge to $x$ we have:
$\map {F_X} {x_n} \to \map {F_X} x$

Let $\sequence {x_n}_{n \mathop \in \N}$ be a monotone sequences with $x_n > x$ for each $n$ that converges to $x$.

Then $\sequence {x_n}_{n \mathop \in \N}$ is a decreasing sequence.

So:

$\sequence {\hointl {-\infty} {x_n} }_{n \mathop \in \N}$ is a decreasing sequence of sets.
$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \hointl {-\infty} x$

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

Note that $P_X$ is a finite measure.

So, from Measure of Limit of Decreasing Sequence of Measurable Sets, we therefore have:

$\ds \map {P_X} {\hointl {-\infty} x} = \lim_{n \mathop \to \infty} \map {P_X} {\hointl {-\infty} {x_n} }$

So we obtain:

 $\ds \map {F_X} x$ $=$ $\ds \map \Pr {X \le x}$ Definition of Cumulative Distribution Function $\ds$ $=$ $\ds \map {P_X} {\hointl {-\infty} x}$ Definition of Probability Distribution $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \map {P_X} {\hointl {-\infty} {x_n} }$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \map \Pr {X \le x_n}$ Definition of Probability Distribution $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \map {F_X} {x_n}$ Definition of Cumulative Distribution Function

So, since $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary:

for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, that converge to $x$ we have:
$\map {F_X} {x_n} \to \map {F_X} x$
$F_X$ is right-continuous at $x$.

$\blacksquare$