# Is the CDF of a continuous random variable continuous?

Let $$X:\Omega\to\mathbb{R}$$ be a continuous random variable in the sense that its range is an uncountable set. Is it possible that its cdf $$F_X:\mathbb{R}\to\mathbb{R}$$ is not a continuous function? I did learn that if $$X$$ is absolutely continuous (i.e., has a density function), then $$F_X$$ is continuous.

Some authors define a continuous random variable to be those whose CDF is continuous, while others define it to be those whose range is an uncountable set. So I guess my question is whether these two definitions are equivalent.

• May be monotone continuity from below and monotone continuity from above of the underlying probability measure will establish continuity of $F_X$? Aug 12, 2020 at 8:38

The definitions are not equivalent.

Let $$\Omega=\mathbb R$$ be equipped with the $$\sigma$$-algebra of Borel subsets and a probability measure $$P$$ that satisfies $$P([0,1])=p>0$$.

Now prescribe $$X:\Omega\to\mathbb R$$ by $$\omega\mapsto0$$ if $$\omega\in[0,1]$$ and $$\omega\mapsto\omega$$ otherwise.

Then clearly the range of $$X$$ is uncountable, but $$P(X=0)=p>0$$.

That means that $$F_X$$ is not continuous at $$0$$.

Let $$X$$ have standard normal distribution. Let $$Y=X$$ if $$X <1$$ and $$2$$ if $$X \geq 1$$. Then $$Y$$ takes uncountably many values. But $$P( Y \leq 2)=1$$ and $$P(Y\leq t) =P(X<1)$$ for $$t <2$$. Hence the CDF of $$Y$$ has jump discontinuity at $$2$$.

Let $$X=0$$ with probability $$1/2$$, and otherwise take values in $$[1/2,1]$$ uniformly randomly. Its law is $$\mathcal L(A) = \frac12\delta_0(A) + |A\cap [1/2,1]|.$$ Its CDF $$P(X is not continuous at zero.