Show that Uniform$(1,5)$ is neither singular nor absolutely continuous with respect to Uniform$(0,3)$.

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Actually, I'm just studying singular continuity, absolute continuity.I know the definitions.And have solved few very basic sums. Now, in this problem, I'm not understanding what does this 'with respect to Uniform$(0,3)$' mean! As I know Uniform$(1,5)$(I think in this question 'continuous' uniform has been meant, not sure though) is always continuous. Could anyone help?

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This is just the definitions. Distribution $\mu$ is absolute continuous wrt distribution $\nu$ means for any (measurable) set $A$, $\nu(A)=0$ implies $\mu(A)=0$. Does that hold for your examples? No, $A=[3,5]$ has measure zero under $Unif[0,3]$ but 1/2 under $Unif[1,5]$. $\mu$ is singular wrt $\nu$ means for any set $A$, $\mu(A)>0$ implies $\nu(A)=0$. Counterexample for your examples is $A=[1,2]$ (or any subinterval). Visually, the first notion means the support of one distribution is contained in the other. The second notion means the support of one notion is disjoint from the other. You can see neither applies to the pair $Unif[0,3]$, $Unif[1,5]$.