Difference between support and set of points at which the density is positive Consider a probability measure $P$ on the Borel sets of $\mathbb{R}^d$ which is absolutely continuous with respect to the Lebesgue measure $\Lambda$, with density $dP/d\Lambda=p$. 
Which is the difference between the support of $P$, say $\text{supp}(P)$ - defined as the smallest closed set $C \subset{\mathbb{R}^d}$ such that $P(C)=1$ - and the set $\{x \in \mathbb{R}^d: p(x)>0\}$?
Are the two sets equal? Or it might be that $ \{x \in \mathbb{R}^d: p(x)>0\} \subset \text{supp}(P)$, with
$\Lambda(\text{supp}(P) \setminus \{x \in \mathbb{R}^d: p(x)>0\})=0$,
i.e. the support and the set on which the density is positive differ by a Lebesgue-null set?
 A: The difference is in the word closed. You can define $supp(P)$ as the closure of the set $\{x \in  \mathbb{R}^d: p(x)>0\}$ but the latter is an open set. The two sets have the same Lebesgue measure so yes they differ by a Lebesgue null set.
Edit: I assumed in my solution that $p$ is a continuous function. If it is not, than $\{x \in  \mathbb{R}^d: p(x)>0\}$ need not be open. You might be able to come up with a sufficiently nasty $p$ so that the closure has bigger Lebesgue measure but I can't think of an example right now.
A: Say $A=\{x:p(x)>0\}$. The other answer is correct in stating that the support of $\mu$ is the closure $\overline A$. But $A$ need not be open, and $\overline A$ need not have the same Lebesgue measure as $A$.
Say $p=\chi_{[0,1]}$, the indicator function of the interval $[0,1]$. Then $A=[0,1]$, which is not open.
Enumerate the rationals as $\Bbb Q=\{r_1,r_2,\dots\}$. Let $I_n=[r_n,r_n+2^{-n}]$ and define $$p=\sum_{n=1}^\infty\chi_{I_n}.$$Then $A=\bigcup I_n$, so $\Lambda(A)<1$, while $\overline A=\Bbb R$, so $\Lambda\left(\overline A\right)=\infty$. 
