check if probability distributions are absolutely continous

Question

Two Probability distributions on $(\mathbb{R}, \mathbb{B})$ are: $P_1(A) = \int_A\beta \exp(-\beta(x-b))\textbf{1}_{[b, \infty]}(x)dx, \beta>0, b>0$ and $P_2(A) = \int_A\beta \exp(-\beta x)\textbf{1}_{[0, \infty]}(x)dx$
Check if $P1 << P2\text{ or }P2<<P1$
I am familiar with definition of absolute continuity but I don't know where to start.

Answer 1

Recall that two measures $\mu,\nu$ are s.t. $\nu \ll \mu$ iff $\mathscr{N}_\mu\subseteq \mathscr{N}_\nu$, that is the $\mu$-null sets are included in the $\nu$-null sets. For families of sets $\mathscr{B},\mathscr{C}$ the statement $\mathscr{B}\subseteq \mathscr{C}$ means that $\forall B \in \mathscr{B},B \in \mathscr{C}$. So, $\mathscr{B}\not\subseteq \mathscr{C}$ if $\exists B \in \mathscr{B},B \notin \mathscr{C}$.

If we take $P_1$, then $P_1([0,b))=0$, so $[0,b)$ is a $P_1$-null set. However, $P_2([0,b))=1-e^{-\beta b}>0$. So, $P_1([0,b))=0\not\Rightarrow P_2([0,b))=0$, so $\mathscr{N}_{P_1}\not\subseteq \mathscr{N}_{P_2}$ and we conclude $P_2\not\ll P_1$.

Now we have: $$\begin{aligned}P_1(A)&=\int_{A\cap [b,\infty)}\beta e^{-\beta(x-b)}dx=\\ &=e^{\beta b}\int_{A\cap [0,\infty)\cap [b,\infty)}\beta e^{-\beta x}dx=\\ &=e^{\beta b}P_2(A\cap [b,\infty))\leq\\ &\leq e^{\beta b}P_2(A)\end{aligned}$$ So, $P_2(A)=0\implies P_1(A)=0$. So, if $A$ is a $P_2$-null set, it is also a $P_1$-null set. Therefore, $\mathscr{N}_{P_2}\subseteq \mathscr{N}_{P_1}$, and we conclude $P_1\ll P_2$. Equivalently, we can invoke Radon-Nikodym: $$P_1(A)=\int_{A}\underbrace{(e^{\beta b}\mathbf{1}_{[b,\infty)}(x))}_{:=f(x)}\underbrace{\mathbf{1}_{[0,\infty)}(x)\beta e^{-\beta x}dx}_{P_2(dx)}$$ so there exists a measurable $f \geq 0$ which acts as a density, so $P_1\ll P_2$.