Singular Random Variable is not Absolutely Continuous
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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 if and only if:
- $P_X$ 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.
$\blacksquare$