Singular Random Variable is not Absolutely Continuous

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

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

Then $X$ is not absolutely continuous.

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

Let $\map \BB \R$ be the Borel $\sigma$-algebra on $\R$.

Let $\lambda$ be the Lebesgue measure for $\struct {\R, \map \BB \R}$.

From the definition of an absolutely continuous random variable, we have that $X$ is absolutely continuous :


 * $P_X$ is absolutely continuous with respect to $\lambda$.

That is:


 * for all $A \in \map \BB \R$ with $\map \lambda A = 0$ we have $\map {P_X} A = 0$.

where $\map \BB \R$ is the Borel $\sigma$-algebra on $\R$.

Since $X$ is singular:


 * there exists $B \in \map \BB \R$ with $\map \lambda B = 0$ and $\map \Pr {X \in B} = 1$.

That is, from the definition of probability distribution:


 * there exists $B \in \map \BB \R$ with $\map \lambda B = 0$ and $\map {P_X} B = 1$.

So:


 * for all $A \in \map \BB \R$ with $\map \lambda A = 0$ we have $\map {P_X} A = 0$.

does not hold and so $X$ is not absolutely continuous.