An exercise in a chapter on Hahn and Lebesgue decompositions goes as follows:
Let $X$ be a random variable, and let $Y = cX$ for some $c > 0$. Let $\mu$ be the distribution of $X$ and $\nu$ be the distribution of $Y$. Then prove or disprove:
- $\mu$ is discrete $\Rightarrow$ $\nu$ is discrete.
- $\mu$ is absolutely continuous $\Rightarrow$ $\nu$ is absolutely continuous.
- $\mu$ is singular continuous $\Rightarrow$ $\nu$ is singular continuous.
For simplicity I assume $\Omega = \mathbb{R}$, the proof shall generalize to other spaces without much trouble.
So, the first one is easy: if $\mu$ is discrete, then there exists a countable set $S$ such that $\mu(S) = 1$, so $\nu(cS) = 1$ (where I abuse the notation by meaning $cS = \{ cx \mid x \in S \}$), so $\nu$ is also discrete.
The second one is a bit harder to do rigorously, but I guess one can resort to the Radon-Nikodym Theorem: since $\mu$ is a.c., it is dominated by $\lambda$: for all Borel $A$ we have $\lambda(A) = 0 \Rightarrow \mu(A) = 0$. But $\lambda(A) = 0 \iff \lambda(A/c) = 0$ (here I abuse notation again setting $A/c = \{ \frac{x}{c} \mid x \in A \}$). We also notice that $\nu(A) = \mu(A/c)$, so $\lambda(A) = 0 \Rightarrow \lambda(A/c) = 0 \Rightarrow \mu(A/c) = 0 \Rightarrow \nu(A) = 0$, so $\nu$ is dominated by $\lambda$, so $\nu$ is a.c.
For the final one we use the Lebesgue decomposition: assume $\nu$ is not s.c., meaning there are non-zero discrete or a.c. components in it. But if we go the other way around from $\nu$ to $\mu$, but this time multiplying by $\frac{1}{c}$ instead of $c$, we get that $\mu$ also contains non-zero discrete or a.c. components, leading to a contradiction!
Does this proof sketch seem reasonable?